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3
THIRD EDITION
JEE
ADVANCED
PHYSICS
Mechanics – I
Rahul Sardana
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Copyright © 2020 Pearson India Education Services Pvt. Ltd
Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128.
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publisher’s prior written consent.
This eBook may or may not include all assets that were part of the print version.
The publisher reserves the right to remove any material in this eBook at any time.
ISBN 978-93-539-4030-0
eISBN: 978-93-539-4426-1
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Registered Office: The HIVE, 3rd Floor, Pillaiyar Koil Street, Jawaharlal Nehru Road,
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website: in.pearson.com, Email: companysecretary.india@pearson.com
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Contents
Chapter Insight
Preface
xix
About the Author
CHAPTER
1
xiv
xx
Mathematical Physics ����������������������������������������������������������� 1.1
General Mathematics: A Review������������������������������������������������������������������������������������������������ 1.1
Algebra ������������������������������������������������������������������������������������������������������������������������������������������ 1.2
Multiplying Powers of a Given Quantity ��������������������������������������������������������������������������������� 1.2
Powers of Ten�������������������������������������������������������������������������������������������������������������������������������� 1.3
Arithmetic Progression (AP)������������������������������������������������������������������������������������������������������� 1.3
Geometric Progression (GP) ������������������������������������������������������������������������������������������������������� 1.4
Coordinate Geometry ������������������������������������������������������������������������������������������������������������������ 1.4
Measurement of Positive and Negative Angles ���������������������������������������������������������������������� 1.7
Factorial������������������������������������������������������������������������������������������������������������������������������������������ 1.9
Series Expansions������������������������������������������������������������������������������������������������������������������������� 1.9
Function: An Introduction ���������������������������������������������������������������������������������������������������������1.10
Representation of a Function������������������������������������������������������������������������������������������������������1.10
Slope of a Line ����������������������������������������������������������������������������������������������������������������������������� 1.11
Concept of Limit of Functions: Meaning of the Symbol x → a��������������������������������������������� 1.11
Derivative of a Function�������������������������������������������������������������������������������������������������������������1.13
Definition of Differential Coefficient ����������������������������������������������������������������������������������������1.14
Mathematical Definition ������������������������������������������������������������������������������������������������������������1.14
Geometrical Interpretation of Derivative ��������������������������������������������������������������������������������1.15
Rules of Differentiation �������������������������������������������������������������������������������������������������������������� 1.15
Important Differential Formulae ����������������������������������������������������������������������������������������������1.16
Applications of Derivative���������������������������������������������������������������������������������������������������������1.20
Increasing and Decreasing Function ���������������������������������������������������������������������������������������� 1.20
Maximum and Minimum Values of a Function ���������������������������������������������������������������������1.21
dy
as Rate Measure���������������������������������������������������������������������������������������������������������������������1.22
dx
Integration: An Introduction������������������������������������������������������������������������������������������������������1.23
Rules for Integration�������������������������������������������������������������������������������������������������������������������1.25
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vi Contents
Definite Integrals��������������������������������������������������������������������������������������������������������������������������1.26
Geometrical Interpretation of Definite Integration ����������������������������������������������������������������1.26
Practice Exercise ����������������������������������������������������������������������������������������������������������������������������1.29
Single Correct Choice Type Questions��������������������������������������������������������������������������������������������������� 1.29
Answer Key–Practice Exercise ������������������������������������������������������������������������������������������������������1.39
CHAPTER
2
Measurements and General Physics������������������������������ 2.1
Scientific Process��������������������������������������������������������������������������������������������������������������������������� 2.1
Observation����������������������������������������������������������������������������������������������������������������������������������� 2.1
Physical Quantity ������������������������������������������������������������������������������������������������������������������������ 2.1
Measurement of a Physical Quantity ���������������������������������������������������������������������������������������� 2.2
Fundamental and Derived Units ���������������������������������������������������������������������������������������������� 2.2
System of Units ���������������������������������������������������������������������������������������������������������������������������� 2.3
Dimensions ����������������������������������������������������������������������������������������������������������������������������������� 2.5
Dimensional Formula������������������������������������������������������������������������������������������������������������������ 2.5
Dimensional Equation����������������������������������������������������������������������������������������������������������������� 2.5
Dimensions of Some Physical Quantities ��������������������������������������������������������������������������������� 2.5
Quantities Having Same Dimensions���������������������������������������������������������������������������������������2.12
Symbols�����������������������������������������������������������������������������������������������������������������������������������������2.13
Principle of Homogeneity and Uses of Dimensional Analysis ��������������������������������������������2.14
Conversion of Units from One System to Another ���������������������������������������������������������������� 2.16
To Derive the New Relations������������������������������������������������������������������������������������������������������2.18
Limitations of Dimensional Analysis���������������������������������������������������������������������������������������2.21
Least Count ����������������������������������������������������������������������������������������������������������������������������������2.23
Significant Figures����������������������������������������������������������������������������������������������������������������������� 2.23
Rounding Off �������������������������������������������������������������������������������������������������������������������������������2.24
Precision and Accuracy of a Measurement �����������������������������������������������������������������������������2.25
Significant Figures in Calculations: Few Examples ��������������������������������������������������������������� 2.25
Order of Magnitude: Revisited �������������������������������������������������������������������������������������������������2.26
Order of Magnitude���������������������������������������������������������������������������������������������������������������������2.26
Errors in a Repeated Measurement ������������������������������������������������������������������������������������������2.26
Mean Value����������������������������������������������������������������������������������������������������������������������������������� 2.27
Standard Deviation (σ ) ��������������������������������������������������������������������������������������������������������������2.27
Standard Error in the Mean �������������������������������������������������������������������������������������������������������2.27
Absolute Errors����������������������������������������������������������������������������������������������������������������������������2.27
Relative and Percentage Error���������������������������������������������������������������������������������������������������2.27
Combination or Propagation of Errors ������������������������������������������������������������������������������������2.28
Vernier Calliper ���������������������������������������������������������������������������������������������������������������������������2.31
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Contents
vii
Concept of Zero Error�����������������������������������������������������������������������������������������������������������������2.32
Calculating Zero Error����������������������������������������������������������������������������������������������������������������2.32
Steps to be Followed While Taking Readings with Vernier Callipers���������������������������������2.33
How to Measure �������������������������������������������������������������������������������������������������������������������������� 2.33
Screw Gauge ��������������������������������������������������������������������������������������������������������������������������������2.34
Construction ��������������������������������������������������������������������������������������������������������������������������������2.34
Pitch of Screw�������������������������������������������������������������������������������������������������������������������������������2.34
Principle of Screw Gauge�����������������������������������������������������������������������������������������������������������2.34
Determination of Pitch of Screw �����������������������������������������������������������������������������������������������2.34
Least Count of Screw ������������������������������������������������������������������������������������������������������������������ 2.35
Determination of Least Count���������������������������������������������������������������������������������������������������2.35
Backlash Error�������������������������������������������������������������������������������������������������������������������������������2.35
Determination of Zero Error������������������������������������������������������������������������������������������������������2.35
Reading a Screw Gauge�������������������������������������������������������������������������������������������������������������� 2.36
Solved Problems �����������������������������������������������������������������������������������������������������������������������������2.39
Practice Exercises���������������������������������������������������������������������������������������������������������������������������2.43
Single Correct Choice Type Questions��������������������������������������������������������������������������������������������������� 2.43
Multiple Correct Choice Type Questions����������������������������������������������������������������������������������������������� 2.54
Reasoning Based Questions ������������������������������������������������������������������������������������������������������������������� 2.56
Linked Comprehension Type Questions�������������������������������������������������������������������������������������������������� 2.58
Matrix Match/Column Match Type Questions������������������������������������������������������������������������������������� 2.61
Integer/Numerical Answer Type Questions ������������������������������������������������������������������������������������������ 2.65
Archive: JEE Main ��������������������������������������������������������������������������������������������������������������������������������� 2.65
Archive: JEE Advanced��������������������������������������������������������������������������������������������������������������������������� 2.68
Answer Keys–Test Your Concepts and Practice Exercises ������������������������������������������������������������2.77
CHAPTER
3
Vectors ��������������������������������������������������������������������������������������� 3.1
Introduction ���������������������������������������������������������������������������������������������������������������������������������� 3.1
Geometrical Definition����������������������������������������������������������������������������������������������������������������� 3.2
Types of Vector ����������������������������������������������������������������������������������������������������������������������������� 3.2
Triangle Law of Vector Addition of Two Vectors �������������������������������������������������������������������� 3.5
Parallelogram Law of Vector Addition of Two Vectors ���������������������������������������������������������� 3.7
Triangle Inequality ����������������������������������������������������������������������������������������������������������������������� 3.9
Polygon Law of Vector Addition ����������������������������������������������������������������������������������������������� 3.9
Properties of Vector Addition����������������������������������������������������������������������������������������������������� 3.9
Multiplication of a Vector by a Scalar���������������������������������������������������������������������������������������3.10
Properties of Multiplication of Vector by a Scalar�������������������������������������������������������������������3.10
Position Vector�����������������������������������������������������������������������������������������������������������������������������3.10
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To Find AB if the Position Vectors of Points A and B are Known���������������������������������������3.10
Subtraction of Vectors����������������������������������������������������������������������������������������������������������������� 3.11
Rectangular Components in 2-D Space ����������������������������������������������������������������������������������� 3.11
Rectangular Components in 3-D Space �����������������������������������������������������������������������������������3.13
Vector Multiplication of 2 Vectors���������������������������������������������������������������������������������������������3.15
Dot Product����������������������������������������������������������������������������������������������������������������������������������3.15
Geometrical Interpretation ��������������������������������������������������������������������������������������������������������3.15
Physical Interpretation����������������������������������������������������������������������������������������������������������������3.16
Cross Product or Vector Product�����������������������������������������������������������������������������������������������3.18
Geometrical Interpretation of Cross Product��������������������������������������������������������������������������3.19
Physical Interpretation����������������������������������������������������������������������������������������������������������������3.19
Directions �������������������������������������������������������������������������������������������������������������������������������������3.21
Lami’s Theorem���������������������������������������������������������������������������������������������������������������������������3.22
Scalar Triple Product (STP) �������������������������������������������������������������������������������������������������������3.22
Geometrical Interpretation of Scalar Triple Product��������������������������������������������������������������3.23
Properties of Scalar Triple Product��������������������������������������������������������������������������������������������3.23
Vector Triple Product (VTP)�������������������������������������������������������������������������������������������������������3.23
Solved Problems �����������������������������������������������������������������������������������������������������������������������������3.25
Practice Exercises���������������������������������������������������������������������������������������������������������������������������3.30
Single Correct Choice Type Questions��������������������������������������������������������������������������������������������������� 3.30
Multiple Correct Choice Type Questions����������������������������������������������������������������������������������������������� 3.39
Reasoning Based Questions ������������������������������������������������������������������������������������������������������������������� 3.40
Linked Comprehension Type Questions�������������������������������������������������������������������������������������������������� 3.41
Matrix Match/Column Match Type Questions������������������������������������������������������������������������������������� 3.43
Integer/Numerical Answer Type Questions ������������������������������������������������������������������������������������������ 3.45
Archive: JEE Main ��������������������������������������������������������������������������������������������������������������������������������� 3.46
Archive: JEE Advanced����������������������������������������������������������������������������������������������������������������������������3.47
Answer Keys–Test Your Concepts and Practice Exercises ������������������������������������������������������������3.48
CHAPTER
4
Kinematics I��������������������������������������������������������������������������������� 4.1
Rectilinear Motion and Motion Under Gravity�������������������������������������������������������������� 4.1
Introduction to Classical Mechanics�������������������������������������������������������������������������������� 4.1
Concept of Point Object (Particle Model)������������������������������������������������������������������������ 4.2
Concept of Reference Frame ��������������������������������������������������������������������������������������������� 4.2
Distance and Displacement (Relative Position Vector)������������������������������������������������� 4.2
Properties of Displacement ����������������������������������������������������������������������������������������������� 4.3
Average Speed and Average Velocity ������������������������������������������������������������������������������ 4.4
Concept of Average Speed ������������������������������������������������������������������������������������������������ 4.4
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Instantaneous Speed and Instantaneous Velocity���������������������������������������������������������� 4.6
Factors Affecting Acceleration of a Body������������������������������������������������������������������������� 4.7
Uniformly Accelerated Motion Systems ������������������������������������������������������������������������4.10
Directions of Vectors in Straight Line Motion ���������������������������������������������������������������4.10
Reaction Time���������������������������������������������������������������������������������������������������������������������4.14
Equations of Motion for Variable Acceleration��������������������������������������������������������������4.17
Graphical Interpretation of Some Quantities ����������������������������������������������������������������4.24
Motion with Uniform Velocity����������������������������������������������������������������������������������������� 4.24
Graphs in Uniformly Accelerated Motion (a ≠ 0)����������������������������������������������������������4.26
Interpretation of Some More Graphs������������������������������������������������������������������������������4.26
Interpretation of Graphs of Various Types of Motion��������������������������������������������������4.32
Graphs���������������������������������������������������������������������������������������������������������������������������������4.34
Vertical Motion Under Gravity���������������������������������������������������������������������������������������� 4.39
Motion in a Plane and Relative Velocity���������������������������������������������������������������������� 4.49
Motion in a Plane: An Introduction ��������������������������������������������������������������������������������4.49
Relative Motion������������������������������������������������������������������������������������������������������������������4.54
Relative Motion in One Dimension �������������������������������������������������������������������������������� 4.55
Relative Acceleration ��������������������������������������������������������������������������������������������������������4.56
Equations of Motion in Relative Velocity Form ������������������������������������������������������������4.56
Relative Motion in Two Dimension ��������������������������������������������������������������������������������4.57
Relative Motion for Bodies Moving Independently�����������������������������������������������������4.58
Relative Motion for Bodies Moving Dependently���������������������������������������������������������4.58
Velocity of Approach/Separation in Two Dimension��������������������������������������������������4.62
Where to Apply the Concept of Relative Motion?���������������������������������������������������������4.64
Category 1: Distance of Closest Approach Between Two Moving Bodies�����������������4.65
Relative Motion in River Flow (One-dimensional Approach)������������������������������������� 4.67
River Problem in One Dimension �����������������������������������������������������������������������������������4.67
Category 2: River-boat Problems or River-swimmer Problems ���������������������������������4.68
Condition for the Swimmer to Cross the River in the Minimum Possible Time �����4.68
Condition for Zero Drift or Condition to Reach the Opposite Point ������������������������� 4.69
Condition When the Boatman Crosses the River Along the Shortest Route ������������4.70
Category 3: Aeroplane-wind Problems ��������������������������������������������������������������������������4.71
Category 4: Rain-man Problems���������������������������������������������������������������������������������������4.73
Solved Problems �������������������������������������������������������������������������������������������������������������������4.78
Practice Exercises �����������������������������������������������������������������������������������������������������������������4.91
Single Correct Choice Type Questions ������������������������������������������������������������������������������������� 4.91
Multiple Correct Choice Type Questions���������������������������������������������������������������������������������4.102
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Reasoning Based Questions������������������������������������������������������������������������������������������������������4.107
Linked Comprehension Type Questions�����������������������������������������������������������������������������������4.108
Matrix Match/Column Match Type Questions������������������������������������������������������������������������4.113
Integer/Numerical Answer Type Questions�����������������������������������������������������������������������������4.119
Archive: JEE Main�������������������������������������������������������������������������������������������������������������������4.122
Archive: JEE Advanced������������������������������������������������������������������������������������������������������������4.125
Answer Keys–Test Your Concepts and Practice Exercises��������������������������������������������������4.129
CHAPTER
5
Kinematics II ������������������������������������������������������������������������������� 5.1
Curvilinear Motion ��������������������������������������������������������������������������������������������������������� 5.1
Introduction������������������������������������������������������������������������������������������������������������������������� 5.1
Angular Displacement, Angular Velocity, Angular and Centripetal Acceleration ��� 5.2
Angular, Centripetal, Tangential and Total Acceleration ��������������������������������������������� 5.5
Unit Vectors along the Radius ( r̂ ) and the Tangent ( t̂ )������������������������������������������������ 5.7
Velocity and Acceleration of Particle in Circular Motion ��������������������������������������������� 5.7
Kinematics of Motion of Particle in a Curved Track ����������������������������������������������������� 5.8
Radius of Curvature������������������������������������������������������������������������������������������������������������ 5.9
Projectile Motion ��������������������������������������������������������������������������������������������������������� 5.12
Projectile Motion: An Introduction����������������������������������������������������������������������������������5.12
Types of Projectile Motion�������������������������������������������������������������������������������������������������5.12
Horizontal Projectile ���������������������������������������������������������������������������������������������������������5.12
Oblique Projectile �������������������������������������������������������������������������������������������������������������� 5.16
Range, Maximum Height and Time of Flight for Complimentary Angles ���������������5.26
Two Unique Times for which Projectile is at Same Height������������������������������������������5.27
Radius of Curvature of an Oblique Projectile at a Point P ������������������������������������������5.29
Equation of Trajectory of an Oblique Projectile in Terms of Range ���������������������������5.30
Relative Motion Between Two Projectiles/Motion of One Projectile as
Seen from Another Projectile �������������������������������������������������������������������������������������5.31
Condition of Collision Between Two Projectiles�����������������������������������������������������������5.32
Motion of a Projectile Up an Inclined Plane�������������������������������������������������������������������5.37
Motion of a Projectile Down an Inclined Plane��������������������������������������������������������������5.38
Solved Problems �������������������������������������������������������������������������������������������������������������������5.45
Practice Exercises ����������������������������������������������������������������������������������������������������������������� 5.53
Single Correct Choice Type Questions ������������������������������������������������������������������������������������� 5.53
Multiple Correct Choice Type Questions���������������������������������������������������������������������������������� 5.66
Reasoning Based Questions������������������������������������������������������������������������������������������������������� 5.69
Linked Comprehension Type Questions������������������������������������������������������������������������������������ 5.70
Matrix Match/Column Match Type Questions������������������������������������������������������������������������� 5.73
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Integer/Numerical Answer Type Questions������������������������������������������������������������������������������ 5.75
Archive: JEE Main�������������������������������������������������������������������������������������������������������������������� 5.77
Archive: JEE Advanced������������������������������������������������������������������������������������������������������������� 5.79
Answer Keys–Test Your Concepts and Practice Exercises����������������������������������������������������5.80
CHAPTER
6
Newton’s Laws of Motion����������������������������������������������������� 6.1
Dynamics: An Introduction����������������������������������������������������������������������������������������������� 6.1
Force ������������������������������������������������������������������������������������������������������������������������������������� 6.1
Newton’s Laws of Motion�������������������������������������������������������������������������������������������������� 6.2
Newton’s First Law or Law of Inertia������������������������������������������������������������������������������ 6.2
Momentum p ���������������������������������������������������������������������������������������������������������������������� 6.2
Newton’s Second Law of Motion������������������������������������������������������������������������������������� 6.3
Newton’s Third Law of Motion���������������������������������������������������������������������������������������� 6.3
Fundamental Forces in Nature ����������������������������������������������������������������������������������������� 6.4
Classification of Forces on the Basis of Contact ������������������������������������������������������������� 6.5
System ���������������������������������������������������������������������������������������������������������������������������������� 6.6
Concept of Impulse and Impulse as Area Under F -t Graph ��������������������������������������� 6.6
Impulse – Momentum Theorem���������������������������������������������������������������������������������������� 6.7
Constrained Motion of Connected Particles�������������������������������������������������������������������� 6.9
One Degree of Freedom����������������������������������������������������������������������������������������������������� 6.9
Two Degrees of Freedom��������������������������������������������������������������������������������������������������6.10
Simple Constraint Motion of Bodies and Particles in Two Dimensions�������������������� 6.11
Wedge Constraints ������������������������������������������������������������������������������������������������������������6.16
Free Body Diagram (FBD)�������������������������������������������������������������������������������������������������6.22
Weight of a Body (W ) ��������������������������������������������������������������������������������������������������������6.22
Normal Reaction/Normal Contact Force ���������������������������������������������������������������������� 6.23
Normal Reaction for Various Situations�������������������������������������������������������������������������6.25
Tension in a Light String���������������������������������������������������������������������������������������������������6.25
Tension in a Rope Having Uniform Mass Distribution �����������������������������������������������6.28
Spring Force������������������������������������������������������������������������������������������������������������������������6.30
Spring Balance��������������������������������������������������������������������������������������������������������������������6.32
Newton’s Second Law: Revisited������������������������������������������������������������������������������������6.33
Atwood’s Machine ������������������������������������������������������������������������������������������������������������6.34
Masses Connected with Strings���������������������������������������������������������������������������������������6.36
Masses on a Smooth Surface in Contact with Each Other�������������������������������������������6.36
Body on a Smooth Inclined Plane �����������������������������������������������������������������������������������6.37
When Masses are Suspended Vertically from a Rigid Support ����������������������������������6.38
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When Two Masses are Attached to a String which Passes Over a Pulley
Attached to the Edge of a Horizontal Table��������������������������������������������������������������6.39
When Two Masses are Attached to a String which Passes Over a Pulley
Attached to the Edge of an Inclined Plane���������������������������������������������������������������6.39
When Two Masses are Attached to a String which Passes Over a Pulley
Attached to the Top of a Double Inclined Plane �����������������������������������������������������6.39
Weighing Machine�������������������������������������������������������������������������������������������������������������6.42
Equilibrium of Coplanar Forces���������������������������������������������������������������������������������������6.46
Rotational Equilibrium (Law of Conservation of Moments of Force) �����������������������6.47
Examples and Situations for Rotational Equilibrium �������������������������������������������������� 6.51
Pseudo Force ��������������������������������������������������������������������������������������������������������������� 6.55
Frames of Reference�����������������������������������������������������������������������������������������������������������6.55
Man in a Lift �����������������������������������������������������������������������������������������������������������������������6.64
Friction�������������������������������������������������������������������������������������������������������������������������� 6.68
Introduction������������������������������������������������������������������������������������������������������������������������6.68
Reasons for Friction�����������������������������������������������������������������������������������������������������������6.69
Contact Force and Friction ����������������������������������������������������������������������������������������������� 6.69
Static and Kinetic Friction�������������������������������������������������������������������������������������������������6.69
Laws of Friction �����������������������������������������������������������������������������������������������������������������6.71
Coefficient of Friction, Limiting Friction and Angle of Friction ���������������������������������6.78
The Coefficient of Friction (μ) ������������������������������������������������������������������������������������������ 6.79
Resultant Force Exerted by a Surface on the Block ������������������������������������������������������6.79
Acceleration of Block on Rough Horizontal Surface����������������������������������������������������6.79
Angle of Repose (α )����������������������������������������������������������������������������������������������������������� 6.79
Acceleration of Block Down a Rough Incline����������������������������������������������������������������6.80
Retardation of Block Moving Up a Rough Incline��������������������������������������������������������6.80
Maximum Height (H ) to which an Insect can Crawl Up a
Rough Hemispherical Bowl ���������������������������������������������������������������������������������������6.80
Maximum Length of Chain that can Hang from the Table Without
Falling from it����������������������������������������������������������������������������������������������������������������6.81
Minimum Force for Motion along Horizontal Surface and its Direction������������������6.81
Dynamics of Circular Motion ��������������������������������������������������������������������������������������� 6.96
Circular Motion: An Introduction �����������������������������������������������������������������������������������6.96
Variables of Circular Motion�������������������������������������������������������������������������������������������� 6.96
Kinematics of Circular Motion ����������������������������������������������������������������������������������������6.97
Relative Angular Velocity�������������������������������������������������������������������������������������������������6.99
Angular Displacement dθ : Revisited���������������������������������������������������������������������������6.102
Angular Velocity ω : Revisited���������������������������������������������������������������������������������������6.102
Angular Acceleration α : Revisited��������������������������������������������������������������������������������6.103
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 12
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Contents
xiii
Radial and Tangential Acceleration ������������������������������������������������������������������������������6.103
Calculation of Centripetal Acceleration ����������������������������������������������������������������������� 6.104
Dynamics of Circular Motion ����������������������������������������������������������������������������������������6.105
Motion of a Particle in a Curved Track and Radius of Curvature����������������������������6.106
Centrifugal Force�������������������������������������������������������������������������������������������������������������� 6.110
Rotor or Death Well��������������������������������������������������������������������������������������������������������� 6.112
Motion of a Cyclist ���������������������������������������������������������������������������������������������������������� 6.114
Circular Turning on Roads��������������������������������������������������������������������������������������������� 6.114
By Friction Only: Vehicle on a Level Road ������������������������������������������������������������������ 6.114
Maximum Velocity for Skidding and Overturning ���������������������������������������������������� 6.115
By Banking of Roads/Tracks������������������������������������������������������������������������������������������ 6.116
By Friction and Banking of Road Both ������������������������������������������������������������������������� 6.117
Conical Pendulum����������������������������������������������������������������������������������������������������������� 6.117
Solved Problems �����������������������������������������������������������������������������������������������������������������6.122
Practice Exercises ��������������������������������������������������������������������������������������������������������������� 6.135
Single Correct Choice Type Questions������������������������������������������������������������������������������������6.135
Multiple Correct Choice Type Questions���������������������������������������������������������������������������������6.164
Reasoning Based Questions������������������������������������������������������������������������������������������������������6.175
Linked Comprehension Type Questions�����������������������������������������������������������������������������������6.176
Matrix Match/Column Match Type Questions������������������������������������������������������������������������6.186
Integer/Numerical Answer Type Questions�����������������������������������������������������������������������������6.191
Archive: JEE Main�������������������������������������������������������������������������������������������������������������������6.196
Archive: JEE Advanced������������������������������������������������������������������������������������������������������������6.200
Answer Keys–Test Your Concepts and Practice Exercises��������������������������������������������������6.206
Hints and Explanations
Chapter 1: Mathematical Physics����������������������������������������������������������������������������������� H.3
Chapter 2: Measurements and General Physics ���������������������������������������������������������� H.5
Chapter 3: Vectors���������������������������������������������������������������������������������������������������������H.41
Chapter 4: Kinematics I�������������������������������������������������������������������������������������������������H.65
Chapter 5: Kinematics II�����������������������������������������������������������������������������������������������H.143
Chapter 6: Newton’s Laws of Motion �������������������������������������������������������������������������H.193
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 13
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2 General4PhysicsKinematics I
CHAPTER
CHAPTER
ChaPter InsIGht
Measurements and
Learning Objectives
Help the students
set an aim to
achieve the major
take-aways from a
particular chapter�
CHAPTER
Learning
Objectives
After reading this
chapter, you will be able to:
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
reading this chapter,
you and
will their
be able
to understand concepts and problems based on:
(a) Physical Quantity and its
(d) Principle After
of Homogeneity
(g) Errors
Propagation
(a) Rest, Motion and
(i) Vertical Motion Under Gravity
(h) Position
Measurements done using
Measurement
and its uses
(b)of Distance
Displacement
( j) Motion in a Plane
(b) Fundamental and Derived (e) Limitations
Dimensional
Vernier Calliper (VC)
(c)
Average
Speed
and
Average
Velocity
(i) Measurements done using(k) Relative Motion in One Dimension
Units
Analysis
(d)
Instantaneous
Speed
and
Instantaneous
(l) Relative Motion in Two Dimensions
(c) Dimensional Analysis
(f) Least Count, Significant
Screw Gauge (SG).
Velocity off
(m) Distance of Closest Approach between
Figures and Rounding
(e) Average and Instantaneous Acceleration
Moving Bodies
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
(f) Uniformly Acceleration Motion
(n) River-Swimmer Problems
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
(g) Variable Accelerated Motion
(o) Aeroplane-Wind Problems
and Advanced) are also given.
(h) Graphical Interpretation and Graphs
(p) Rain-Man-Wind Problems
6
Newton’s Laws of
Motion
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
Learning
Objectives
After reading this
chapter, you will be able to:
and Advanced) are also given.
scientific Process
Objective Observation
After reading this chapter, you will be able to understand concepts and problems based on:
An
observation that remains identical for all
The history(a)
of Newton’s
science reveals
that it has evolved
Laws of Motion
(c) Friction
3.8
Advanced
JEE
Physics
the
observers
(persons)
is called an Objective
of mechanics depends on
through
a
series
of
steps.
Let
us
have
a
small
discus(b)
Pseudo
Force
(d)
Dynamics
Circular
Motion
so, to conclude we have
RECTILINEAR
MOTIONofAND
MOTION
uNDER GRAvITY
Observation.
ical quantities all having
sion of the steps
involved.
All
this
is
followed
by
a
variety
of
Exercise
Sets
(fully
solved)
which
contain
questions
as per the
1. sin x , cos x , tan x , cosec x , sec x , cot x , log x , e x , a x
, then dimensional analySince
it
is
given
that
n
a about
unit vector,
therefore
EXAMPLE:
latest
JEE
pattern.
At
the
end
of
Exercise
Sets,
a
collection
of
problems
asked
previously
in also
JEE (Main
1
s is
relativistic
mechanics is that it is mo
STEP-1: Observation -1 ⎛ ⎞
INTRODuCTION
TO
CLAssICAL
⇒ θ = tan ⎜ ⎟
all are dimensionless.
erive their relationship.
we viewing
have the same painting mayand
Four observers
feelreduces
dif- to classical mechanics for the ca
and Advanced)
⎝ 2are
⎠ also given.
STEP-2: Proposing/propounding
a theory MEChANICs
based on
ness does not establish
i.e., [ sin x ] = [ cos x ] = [ tan x ] = [ cosec x ] =
ferently
about
the
“beauty”
of
the
painting,
where
as
(non relativistic) velocities. In this chapter,
those observations.
1 = 1 + 1 + 2 cos θ
but reverse is true.
shalldealing
report identical
length,
breadth or area
of the and studying motion in terms of
branch ofthey
Physics
with motion
of particles
describing
Verification of the theory as appliedThe
to those
[ sec x ] = [ log x ] = [ e x ] = [ a x ] = M0L0STEP-3:
T0
1
painting,
when
asked.
So,
Beauty
is
not
an
objective
or
bodies
in
space
and
time
is
called
Mechanics.
As
time while ignoring the causes that produ
observations.
⇒
cos
θ
=
2. The argument of all the functions i.e. x is also
observation
it cannot
2 be assigned
long as the velocity
of theasmoving
bodies
is small a numerical
This particular part of Classical Mechanic
STEP-4: Modification in the5theory,+ifB at all necessary.
l Note(s)
A
dimensionless. Hence
motion
anythe
object.
Force
is the
inter- Furthermore, in this part o
value
along
with
appropriate
unit
(that
could
have
dYNAMIcS: AN INTrOducTION
in comparison
todirection
the
velocity
light
linear
Kinematics.
⇒of some
θ of
= 120
°of(c),
action
between
theremain
object invariable
and the source
θ
measured
it).
dimensions and
the
time
intervals
[ x ] = M0L0 T 0
we(providing
shall be limiting ourselves to the mot
obserVation
q, cosq, tanq (and their
Theory
with
In kinematics, we studied the motion
of
a
particle,
10
The
difference
the two
vectors
is SI
given
the
pull
or(platform(s)
push).
It isobjective
aof
vector
quantity.
Its
unitby
is
Physics
always
deals
with
observation.
in all Reference
Frames
from
where
dimension
and two dimensions i.e., moti
is dimensionless and
with emphasis
on motion
along
a straight
lineisand
cgs unit
(N)i.e.,
and
is dyne
Observations
are basically
of two
types
motion
being newton
observed),
not
depend
straight line, also called as Rectilinear M
nd they
= nˆ 1 -do
nˆ 2 Illustrations
motion
in
a
plane
and
simply
described
it
in
terms
illustration 13
5
is also dimensionless.
on choice of reference
dealing with
planar motion. From our everyday expe
illuStRAtion
7
1frame.
10Mechanics
N =Quantity
dyne
Physical
r , v and a. Now comes thesuch
timelike
when
of vectorsObservation
Subjective
( 120
nd2 =asn12Non-Relativistic
+ n22Elaborative
- 2n1n2 cos θ = 1observe
+ 1and
- 2 cos
°)
motion (also ⇒
called
that
actually
motion represents a c
In
figure,
a
particle
is
moving
in
a
circle
of
radius
Chapter
3:
Vectors
3.13
A
force
acting
on
a
body
If
V
is
velocity,
F
is
force,
t
is
time
and
we shall be discussing about the cause
producing
are also dimensionless.
The objective
quantities
to which
a numerical
value
motion)
is called
as Classical
Mechanics.
However,
change
in the position of an object. In Phys
An observation
that at
varies
fromconstant
person to
person
isWhat
simple
theory
r
centred
O
with
speed
v
.
is
the
1
⎛
⎞
F
2
motion. This treatment, happens to be when
an aspect
of can (a)
... ) is dimensionless and
be
attached
ed motion
to
the bodies
move
with
speeds
into three categories
nalong
2 -with
2comparable
+ to
1 =(specifi
3 divide
may⇒
change
speed.
called Subjective
Physics never
deals
⎜⎝ - some
⎟ = 2unit
α = sin ( βt ) , then find the dimensional formula
d =only
A
change
inObservation.
velocity
in moving
from
to B
?measure
Given
2 ⎠Quantities.
mechanics,
known
as dynamics.
In this
chapter,
mensionless. So,
it) are
called
Physical
Else,
they
V2 speed we
of light
(called
as
relativistic
speeds),
then
helps
the
students
(b)
may
change
only
direction
of
motion.
with subjective observations
like beauty, the
emotion,
⇒ OA = -2 2iˆ - 2 2 ˆj
71
(a)
Translational
Motion (studying now).
∠
AOB
=
60
°
.
shall be primarily
along
of α and
may(c)
also
be⇒
defi
asboth
the
quantities
which of
tan a = discussing the forcesthe
part with
of Physics
dealing
with
such
like
may
change
themotion(s)
speedthrough
and direction
personality etc.
β .
nned
d = 3
94 and properties that
EXAMPLE: A car moving down a highwa
to understand
their respective nature
for
isaccount
called Relativistic
Mechanics. An interesting
fact
iˆ AB = 5 cos 0°iˆ = 5iˆ
motion.
solution
the motion
As done before, the bodies will
⇒ aof= a37body.
°
(d) may change size andthe
shape
of a body.
illustrations
iˆ BC = 6 cos 60°iˆF + 6 sin 60° ˆj = 3iˆ + 3 3 ˆj
illuStRAtion 9
be treated as if they were single particles. However, in
α =2 sin( β tt)
So resultant
iswe
118
N atbe180
- 37 = 143
°. proper
the
the
later
chapters,
shall
extending
the
V
supporting
imensionless
02_Measurements, General Physics_Part 1.indd 1
11/26/2019 5:02:08 PM
⇒ OC = OA + AB + BC
6.12 Advanced JEE Physics
Unit ofConsider
Force two vectors
A and B inclined
at an angle
O
ties of this chapter to discuss the 04_Kinematics
motion of
a group
1_Part
1.indd 1 of
V
ˆ
ˆ
ˆ
ˆ
ˆ
-2
B
theory�
Please
note
θ
.
If
A
=
B
=
R
,
then
prove
that
-2 2i - 2 2 j + 5i +dimensionless
3i + 3 3 j
⇒ OC = dimensionless
SI
unit
of
force
is
newton
(N)
and
1
N
=
1
kgms
particles
and
extended
bodies
as
well.
So,
this
branch
RECtAnGulAR CoMPonEntS
in
60°
-2
B with the cause
get
ILLuSTrATION 9
gcms
cgs
unit
of
force
is
dyne
(dyn)
and
1
dyne
=
1
Differentiating with respect to t, we
of
Physics
dealing
with
motion
along
3-D SPACE
that
⎛ θ ⎞theory and
⇒ OC = ( -2 2 + 5 + 3 ) iˆ + -2 2 + 3 3 ˆj
60°
A +=B10=5 2dyne
R cos. ⎜ ⎟
producing
is calledrod.
Dynamics.
⇒ [ βt ] = M 0 L0 T 0 Also, 1 (a)
newton
VA
Figure shows a hemisphere
and motion
a supported
⎝ 2⎠
dy
dx
A
problem
solving
In
3-D
space,
we
have
ˆ
ˆ
dimensionless
+ 2y
=0
2x
-2 ] of Motion 6.11
⇒ OC = +5 ⋅ 17 i + 2.37 j
Hemisphere is moving in right direction with a uni[
Chapter
6: Newton’s
Laws
The dimensional
of force
dt
dt
formula
⇒ [ β ] = T -1 θ ⎞ is MLT
⎛
FOrcE
form velocity v2 and the end of rod which is in contact
(b)
A
B
=
2
R
sin
techniques
areis kilo⎜
⎟
another
commonly
used
unit
of
force
R
=
R
+
R
+
R
⎝
⎠
xv + yvB = 0
Solution
x
y
z
2
[
]
F
with ground is moving in
left direction
a velocgram force (kgf). It is the force with which a body
illuStRAtion
[ α ] =12
Force
is
a pull with
which
or tends
Similarly
changes
IllustratIon
IllustratI
Illustrat
Ior
on push
7
later on the developed
can be linked for getbasedrelations
on simple
Change
Δ
= vuniform
ity v1. Find the rate at which
the angle
qvelocity
is changing
[V 2 ]
of massSolution
1 kg is attracted towards the centre of the
B - vA
⎛ x⎞
Rthe
= in
Rstate
+ R=zor
k v of
ofy jrest
motion or
xi+ R
Four coplanar forces act on a body
at point O asto change
ting the required parameters. Sometime x and y direc⇒ vB = - ⎜ ⎟ v
The car
A is used to pull a load B with the pulley
in
terms
of
v
,
v
,
R
and
q
.
learning
program
1
2
1
1
2
⎝ y⎠
or any two directions of the motion are
imensionless
shown in[ M
diagram
of rectangular component
tional motion
L T ] by use
Since
vB - vA shown.
= vB2 +IfvA2A-has
2vA vaBforward
cos θ
arrangement
velocity vA,
(a) Since A = B = 1
⇒ direction
α=
= M1 L-1T 0
y
find
related by someIF
specific
rule, we →
call such rules as
-1 ]2 magnitude of resultant force.
→2 THEN
determine
an expression for the upward velocity vB
⇒ vB = -v cot q
[ L1T and
constraint
⇒
- vAin=terms
v 2 +of
v 2x.- 2v 2 cos ( 60° )
⇒
Arules.
+ B =These
A +rules
B2 + 2relate
AB cosone
θ direction of
of
thevBload
R
v
100 N 1
ELSE�
would
v2
{negative sign indicates, y decreasing with time}
motion of an object
with I
some
other direction of the
θ
illustration 14 110 N
2
2
2
⎛ 1⎞
A +orBsome
= R other
+ R +object
2you
R cos
θ = 2R 1 + cos θ
same⇒object
also.
suggest
not
⇒ 1.indd
vB -1vA = vR2 y+ v 2R- 2v 2 ⎜ ⎟ = v
x of Motion_Part
06_Newtons Laws
11/26/2019 12:03:22 PM
mETHOD II
dimensionless
⎝ ⎠
FV 2
⎛ 2πβ ⎞
Rx 2
In cases when the relation between
points
of
a
2⎛θ⎞
log
where
If, α = two
F
is
force,
V
is
x
⎜
⎟
e
2
Since 1 + cos
θ = attempt
2 cos ⎜ ⎟ the
IllustratIon
8 to
45°
⎝ V 2 30°
⎠
β
⎝ 2⎠
x
rigid body is required, we can make use of the fact
l
SOLuTION
O
illuStRAtion
Rz
80
N
zh 8
Figure shows a illustrations
rod of length l resting on a wall and
velocity,
then160
find
the20°
dimensional
formula
of α and
that in a rigid body the distance
between
two
N points
⎛
⎞
Here
x
is
the
separation
between
centre
of
hemisphere
the floor.
lower
A ⎜isθpulled
towards left with a
. relative velocity of
If the sum of two
unit vectors is a unit vector, then
⇒ Its
A+
B = 2end
R cos
always remains same. Thus βthe
B
R makesxan
a with x axis, b with y axis and
⎝ 2 ⎟⎠going
without
and the end of rod. Rate of Ifchange
is angle
actually
constant velocity
u. Find the
velocity of the other end
find the of
magnitude
of difference of these
two vectors.
one point of an object with respect to any other point
imensionless
A
g of
with
z axis,
then
solution
Solution
the
relative
velocity
of
end
rod
and
centre
of
B downward
the rod makes
angle q with the
of the same object in the direction of line joining them
(b)
Similarly, when
through
the an
theory
x
are required to find the
horizontal. ( v1 + v2 ). We Solution
2
2 πβhemisphere
and Fv
their
components
are asi.e.,
follows
will always remain zero, as The
theirvectors
separation
always
2
2
2
loge
α= 2
A
B
=
R
+
R
2
R
cos
θ
= 2R 1 - cos θ
R
R
R
of
that
section�
2
z sum
y
Let
n̂1 =andx n̂
two unit
vectors,
V
β
dq
dx2 are the
remains constant.
cos a
costhen
g = the
solutIon
= v1 cos
+ vb =
rate of change of q, i.e.,
, knowing
that
Magnitude
R dt of
R
B
and difference
these2 twoRvectors be represented
by
Here in above example the distance between the
θ
dt
⎛ ⎞
of
x component of
y component of
Since 1 - cos θ = 2 sin 2 ⎜ ⎟
We designate
the position
of the car by the coordinate
imensionless
points A and B of the rod always
remains
constant,
⎝ 2⎠
dimensionless
dimensionless
resultant
resultant vector
Since,
xresultant
= R cosecvector
q
ˆ2
ns the
= nˆ 1position
+ nˆ 2 andofndthe
= nˆload
1 -n
x
and
by
the
coordinate
y,
thus, the two points must have
same
velocity
vector
Rx
Rx
Differentiating with respect to⇒
time
we
get
⎛ θ ⎞l
2 a =2
=2from a fixedθ reference.
bothcos
The
components in the direction of their line joining i.e.,
⇒
nmeasured
=2 1=+l1 + 2 cos
θ total conA - B = 2R sin ⎜ ⎟
⇒
2
s = n1 +
R n2 + 2n21n+2Rcos
⎝ 2⎠
80
80
0
cable
is
x
y + Rz
dx
dq stant length of the R
along the length of the rod.
= - R cosec q cot q
If point B is moving down with
velocity110
vBcos
, its45° = 71
dt 100 sin 45° = 71
dt
A
2
2
θ
100
v
L = 2 ( hR−
y y ) + l = 2R(yh − y ) + h + x
=m
componentF01_Physics
along theforlength
of the
rod is vB sin q .
JEE Mains
and Advanced_Mechanics
I_Prelims.indd 14
11/28/2019 7:53:51 PM
d ⇒ cos b = R =
2
+ R2 + R2
Rq
∵ ( cosec q ) = - cosec
q cot
(
)
(
(
)
)
For n moles of gas, Vander Waals equation is
a ⎞
⎛
P + 2 ⎟ ( V − b ) = nRT . Find the dimensions of a
⎝⎜
V ⎠
and b , where P is gas pressure. V is volume of gas
and TAdvanced
is temperature
of gas.
3.14
JEE Physics
Since [ Fv ] = M L T
⎡ β ⎤
So, ⎢ 2 ⎥ should also be M1 L2 T −3
⎣x ⎦
⇒
⇒
SOLUTION
a ⎞ eptual Note(s)
⎛C o n
P+ 2c
− b) = nRT
⎟ × ( V
⎝⎜
V ⎠ volume
pressure
(a) If l, m, n are called Direction Cosines of the vector,
then l 2⎡ +am⎤2 + n2 = 1
⇒ [P] = ⎢ 2 ⎥ and [b] = [V ] = L3
⎣ V2 a ⎦+ cos2 b + cos2 g = 1
cos
[ β ] = M1L2T −3
[ x2 ]
[ β ] = M1L4 T −3
Solution
β ⎤
⎡
1 2 −3
and ⎢ Fvvector
+ 2 ⎥ vwill
also angle
have dimension
L T x-,
Velocity
makes
a , b and M
g with
x
⎣
⎦
y- and z- axis respectively
[α ]
1 2 −3
LT
∴⇒ a =2 60=° M
and
g = 60°
[t ]
2
2
2
Since, cos a 1+2cos
−1 b + cos g = 1
[
]
⇒ α = M LT
2
2
2
⇒ cos 60° + cos b + cos 60° = 1
Chapter Insight
xv
Test Your Concepts
(b) In 3-D space a vector of magnitude r making an
2
2
angle
a withConcepts-I
x-axis, b with y-axis and g with z-axis
Test Your
⎛ 1⎞
⎛ 1⎞
2
These topic based
⇒ ⎜ ⎟ + cos b + ⎜ ⎟ = 1
can thus be written as
⎝ 2 ⎠ and Verifi⎝cation
2⎠
Based on Principle of Homogeneity
2.30 Advanced JEE Physics
exercise sets are
1
iˆ r = r ⎡⎣ ( cos a ) iˆ + ( cos b ) ˆj + ( cos g ) kˆ ⎤⎦
(Solutions on page H.5)
⇒ cos 2 b = 1 - 4
based on simple,
1. If a composite physical quantity in terms of
2
π Pr
V=
l
m
moment
ofDA
inertia
18aηll N o t e ( s )
DB F,
single concept
⎞ m I,⎛ force
⎞ velocity v, work W C o
n
c
e
p
t
u
l⎛
⇒ cos b =
A ⎜ 1±
⎟⎠ B ⎜⎝ 1 ±
⎟
2
⎝
DX ⎞
2
⎛
A
B ⎠Q = ⎛ IFv ⎞ . Find the
Check
the
dimensional
consistency
of
this
and
L is definedn as,
⇒ X ⎜ 1 ±illuStRAtion
= length
classification
⎟
13
⎜
⎟
±
N
⎝
ˆ is any
X ⎠
⎝ WL3 ⎠
And if XSince
= kA
constant
andg kN
DC ⎞
⎛
b ˆj + v cos
iˆ equation.
v ,= where
v cos akiˆ +isvacos
-1
C nof⎜velocity
1±
A bird dimensions
moves with
technique� These
⎝ Q. C ⎠⎟ of 20 ms in the direcreal number.
5. Check
1the correctness
1 ˆ
1of the equation:
tion2.making
angle
60° quantities
with eastern
anddimen60°
Can two
physical
haveline
same
× ( ωiˆ t++20
×, where
j + 20 ×y kˆ = displacement,
DX⇒ vDy=
A= 20
a sin
ϕ
)
l
m
are meant for
2
2
2
Then
=N
with vertically
upward.
sions? Explain
DARepresent
DB the
⎛ with
⎞example.
⎛
⎞ velocity vector
A= amplitude, ω = angular frequency and ϕ is an
1±
⎜⎝ 1 ±
⎟⎠ ⎜⎝ universal
⎟⎠ gas constant R, X
a
D
Bmform.
3.X ⎞FindAldimensions
of
ˆ
ˆ
ˆ
⎛ in rectangular
A
B
students practice
⇒ vangle.
= 10i + 10 2 j + 10 k
⇒ X ⎜ 1±
⎟=
DX
n
⎛ DA ⎞
⎝
G.
Chapter 5: Kinematics II 5.15
X ⎠universal
C n gravitational
C⎞
Dconstant
% 6.
⇒
= NIf⎜ E, M,
⎛
⎟⎠ ×J100
and%G respectively denote energy, mass,
after they
study
⎝
1
±
⎜
⎟
X
A
4. The rate of flow⎝(V) ofC a⎠liquid flowing through a
angular momentum and gravitational constant,
P
a
particular
topic
⎛
⎞
2
Test Your radius
Concepts-I
l
-n
rDB
andm a pressure
⎟ is
Test Your
Concepts-IIof EJ .
DX ⎞ ⎛pipe Dof
A⎞ ⎛
DC ⎞ gradient ⎜⎝ illustration
⎛
⎞ ⎛
⎠
calculate
the dimensions
16
⇒ ⎜ 1±
= ⎜ 1±
1±
1Based
±
…(1)
5 2
⎟
⎟
⎜
⎟
⎜
⎟
and
want
to
on
Addition,
Subtraction
and
Resolution
MG
⎝
⎝
⎝
X ⎠ ⎝givenAby⎠Poiseuille’s
B ⎠ equation:
C ⎠
Based on Horizontal Projectile
If V = ( 50 ± 2 ) V and I = ( 5 ± 0.1 ) A, then find the
practice(Solutions
more on
onpage H.134)
(Solutions on page H.31)
DA
DB
DC
percentage error in measuring the resistance. Also
Since
1.1 , Canthree
1 andvectors not
1 , soinfrom
oneBinomial
plane give a zero
6. Establish the following vector inequalities:
1. Awith
body
is thrown
horizontally from the top of a
Calculate the
horizontal
launch
velocity of the
find the resistance
limits
of
error.
that
topic
learnt�
A
B
C
resultant? Can four vectors do?
(a) tower
a - b and
≤ a strikes
+ b the ground after three seconds at
Theorem, we have
. case
rock. TakeFinally,
g = 10 ms -2 in
solution
2. The x and y components of vector A are 4 m
and
an
of
angle
45° with the horizontal. Find the height
l
6. Calculate the minimum velocity u along the hori(b) of
a -the
b ≥tower
a - and
b the speed with which the
6 mPhysics_Part
respectively.
DA ⎞General
DA 1.indd 15The x and y components of vector
⎛
body5:39:15 PM
02_Measurements,
11/22/2019
any
diffi
zontal suchof
that
the ball
justculty
clears the point C.
⎜⎝ 1 ±
⎟⎠ ≅ 1 ± l
(a) Given V = ( 50 ± 2 ) V and I = ( 5 ± 0.1 ) V -2
When
does
the
equality
sign
apply?
was
projected.
Take
g
=
9
.
8
ms
.
A
A
A
+
B
are
10
m
and
9
m
respectively.
Calculate
for
Assume
that
the
ball
is
launched
by a man who
they
can
refer
V
7.
Find
the
magnitude
and
direction
of
the
resultant
mthe vector B the following:- n
2. With
We know that
R = what minimum horizontal velocity u can
holds the ball at a distance 1 m above A. Also
DB ⎞
DB
DC ⎞
DC
⎛
⎛
I
of
following
forces
acting
on
a
particle:
1±
≅ 1 ±itsmx andand
1±
y components,
to the
hints
a boy throw a ball at A and have it just clear the
⎟ (a)
⎟ ≅ 1∓ n
find x, where
the ball
strikesand
the ground. Take
⎝⎜
⎝⎜
B ⎠
B
C ⎠
C
3obstruction
2 DKgf
F2 = 6 2 Kgf due
2
DR F1 =DV
I due
.1
2 at 0north-east,
(b) its length and
B?
.
g = 9.8 ms -solutions
⇒
=
± and
= F += 2 Kgf due north-west.
to
these
south-east
So, (1) becomes(c) the angle its makes with x-axis.
R
V
I
503 5
u
40 m ˆ
ˆ
ˆ
ˆ
exercise
sets given
1m
the sca8.
DR What does2the statement
0.u1 a + b = a + b imply?
DX 3. ⎛ If AD=A4⎞i ⎛- 3 j and
DB ⎞B⎛ = 6i +D8Cj,⎞ obtain
A
in % ) =
×
100 +
× 100 = 6 %
⇒
(
1±
= ⎜ 1± l
1± m
1∓ n
⎟
⎜
⎟
⎜
⎟
A
(
)
9.R Two forces50F1 and F2 acting
at a point have a resul⎝ lar magnitude
⎠⎝
⎠directions
⎝
⎠ A , B, A + B ,
5
and
of
at
the
end
of the
X
A
B
C
3.9 m
tant R1. If F2 is doubled, the new resultant R2 is at
( A - B ) and ( B - A )⎛. Neglected ⎞
V 50
C
B
book�
DX
DA
DB
DC
(b) R = = right
= 10
ohm
angles
B F2 have the
36 mto F1. Prove that R1 and
⇒ ±
6m
=4.± l A particle
±m
+
∓ n a displacement
has
of⎟ 12 m towards
I
5
same magnitude.
X
A
B
C ⎜⎝ Terms
⎠
16.4 m
east and 5 m towards north and 6 m verticallyDR DV DI
0
.
1
2
10. In figure,
5.6 of radius
= a +particle is moving in a circle
upwards.
DX
D
DBFind the
DC magnitude of the sum of these R = V r +centred
A
Advais the
I
50at O5with constant speed v. What
nced
⇒ ±
= ± l displacements.
±m
∓n
14.7 m
JE P
X
A
B
C change
invelocity
velocity
inthe
moving
from
A to B? Given
hysic
3.
The
of
water
jet
discharging
from Ethe
.
2
0
1
s
5. Show that if two vectors are equal in magnitude,⇒ DR =
⇒ .
×
10
+
×
10
=
0
6
Since, we know that during propagation errors are
aC =be obtained
∠AOB
= 40
.5
ce °(small
hole in the tank) can
50 orifi
-w 2r
their vector sum and difference are at right angles.
always taken to be the extremum, so we have
D
from
u0=.8 )2ohm
ghChapter
, where 3:hVectors
= 5⇒m isa3.9
the depth of the
(
⇒
R
±
D
R
=
10
±
C = a
DB
DC
DX
DA
C = rw 2 the
x
orifi
ce
from
the
free
water
surface.
Determine
P
l
ease
+
= +l
+m
+n
not the orifice to
X
A
B
C
tripeleaving
time for Identify
a particle of water
(c) Existence of additive
Solving
Technique(s)
tRiAnGlE
inEQuAlitY
tal a e that fr Problem7.
Ball bearings
of diameter
2v v
…(5 20 mm leave the horizon point B and the horizontal
om
ccele x, where
t
reach
h
Problem
solving
technique(s)
)
e
vav = 1 2
OR
For every vector a , we have
(5),observetalthat
nega distance
r
we
{∵ of magnitude
with a velocity
ILLU u and fall through
we c
v
t-iv2 .e ationSo,
Since a vector cannot have a resultant more than the
1 + v2
it
hits
the
surface.
Take
g
=
10
ms
S
i
r
s
T
oncl
sign
RAT
direc
= r }hole at a depth
(a) Dy ais+always
i.e. Dy > 0.
theu60
mm diameter
ofIO800 mm as
just(a)
0 = 0 +positive
a
DB
DC
DX value
DA
deis
tean
A time
maximum
d rinterval
n equal
thdivided
ifyiwhen
solidparts, N 3 (h) Similarly, when a p
Also
= -l and
- mless -than
n the least value of the
ashown.
at th into
same as that of y.
d
n
,
i
g
a
Calculate
the
permissible
range
of
u
which
if
(b)
Dy
has
units
boof
lly inspeed eiscthe
ththe
we a )
X so1.indd
C
tionaverage
resultant,
we14A
have B
is “average
dy r
i.e., Adding 0 i.e., null vector to any vector
then
en- simple
03_Vectors_Part
cent (say
wathe
ary a the
otatTake
ripet had bee 11/7/2019 3:14:16 PM ). will enable
rdsball
5absolute
m
A is
bearings
es wtance at speed v1, n
xis w hole.
wheto enter
(c)does
A quantity
in terms
of magnitude
error
n ask the speeds in the respective
al foas
(seintervals”.
not changes
the
ofhaavueexpressed
as well
FromRboth≤these
relations, we get
e
ith dthe last one thir
r
ith a conr
e
e
c
and
R
≤
R
d
e
ju
w
the
dotted
positions
to
represent
the
limiting
to giv an interval is divided into cnonequal length
and (b) when
min
max
ecele
n
asdirection, because 0 has zero magnitude
i
s
an
mass st mand
its
t
its a
an
e the
ltipmli
-2 stant.
of th u2.5
ngul gular de rat3iovnv av
DX
ed th genparts,
e
t
a
F
ditions.
Take
g
=
10
ms
ia
x
v
⇒ A
- B= ≤l D
RA≤ +Am+ D
BB + n DC
i
p
e
n
l
e
a
then
the
“reciprocal
of
the
average
speed
r
B
d
=
±
D
units
y
y
y
force
rage
has an arbitrary/indeterminate
direction. partic
r ve v =celera 1 2 b3
essio
(
)
e
t
t
h
a
e
b
l
av
d
F
,
ocity v v +tiovnv +
ns fo
le, sa
o
mea
X
A
B C
tial m
ovethe
c = xma
en average
of
r the reciprocals
e toththe
na
y m, isveequal
rt
(d) Existence of additive Inverse
an1 2 2 3a
e wo of
If A > B , then A - B ≤ A + B ≤ A + B
c = m (
to ge xpressio w20
(d) If least count is notgiven and a measurement
is w0 . oment o he whol ngular v d k is
uld
ns w mm
speeds
intervals”.
t in respective
For a given vector a , there exists a vector ( - aw) × v )=
f tim
Solving
e tim
a
⎛Problem
⎞
⎛ %age ⎞
eloci
120
mm
%age
⎞ ⎛ %age
⎛ %age ⎞
it
h
2
given, then
error
in the measurement
willvelocity
be ±1 u of a- w
e its
e of
ty of
4.
Determine
the
horizontal
tennis
t
h
e
r
(c)
Time
Average
Speed:
When
particle
moves
with
angu
ro
⎜ Error ⎟ + n ⎜ Error ⎟
- B ⎟≤=Al ⎜+Error
B ≤ ⎟A+ +m B
⇒ ⎜AError
such that
anAdvanced
SOLU u
ProAbat height of 2.25 m from
d F = JEE Physics
6.16the
lar v ta3tion i th
in last
digit.
Techniques
ball
launched
from
T
⎜⎝ in y ⎟⎠ ⎜⎝ in
⎟
⎜
⎟
ILLuSTRATION
T
I
m
l
A ⎟⎠ ⎜⎝ in O
f
e
different
uniform
speed
v
,
v
,
v
…
etc.
in
different
e
B ⎠
aT =
N
a
m So
⎝ in C ⎠
locit
1 2 3
a + ( - a ) = ground
0
m( y wa t t
A - B ≤ A+B & A+B ≤ A + B
lvinet
so that it(ajust
the
of height 1 m
) Foclears
ng T
time intervals t1, at2×, tr3,)… etc. respectively,
s esp
Calculate the average
r mo
a = its averqu
echnA. Also,
These techniques
The vector ( - aat) B,is acalled
the additive
inverse
of from
t
horizontal
distance
of
6.4
m
k
io
(
)
i
n
⇒
v
+
4
5
=
0
c
w
que(age
speed over the total time
is given
an u
{called Triangle Inequality}
with
800 mm in the following cases m
⇒ of journey
se th from
s)B
dw
unifnet,
and
vice
versa.
a
fi
nd
the
horizontal
distance
the
where
e
o
ensure that
rm
foll
-2⇒
ang as
CASE-1: For a train tha
v = -20 ms -1 = 20 ms -1 , upwards = k w
ingms
w = wTake go=w10
the ball strikes the ground.
eq . ulaBr acc Total distance covered w dt
a
0 +from
tions vav e=lera
PolYGon
lAW oF
VECtoR ADDition
a t top of uaatower
students
become
another at a uniformA spe
5. A rock is thrown horizontally
tion, time elapsed
⇒
Problem Solving technique(s) q =
d
Total
t
w
w
e
to first station at a speed
11/11/2019
4:43:47
PM from
02_Measurements, General Physics_Part 1.indd 30
and hits the ground 4 s later.
The
line
of
sight
w
1
60 mm
capable
enough
to are represented by
0t +
=
If a number
of non
zero vectors
2
a tthe
(a) If the vectors form
sided
with
CASE-2: For a man who
v2 t2 +-v3kt3dt+ ......
top aofclosed
towernto
thepolygon
point where
rock hits the v = d1 + d2 + d3 + ...... = v1wt01 + w
2 all
w 2resultant
Problem
the (solve
n - 1 ) sides
of an n -sided
av Solving Technique(s)
-of
for the first one minute
the sides in theground
same order
then
is 0. the horizontal.
0 + ......
a variety
of polygon then the
w0230°
makes
an the
angle
with
⇒
+
+
+
+
+
t
t
t
......
t
t
t
= 2a
1
2
3
1
2
3
(
w
2
q
resultant
3 ms -1 for the next oneAm
w ) we can have
For pulleys interconnected with strings,
⎛w+
q –C
problems in an easy
w
=
=
w0 ⎞
- kparticle
0
C
(d)
Distance
Average
Speed:
When
a
⎜
CASE-3:
For a man who
the
following
methods
to
analyse
constraint
equations.
t
⇒
⎝
B
B
D
D
C
-1
wher
2 ⎠⎟ t
describes different distancesw0d1-, d2, d3, … with difspeed of 2ams
thenof
and quick manner�
B = 2 (, Acc.
e
w = kt
(a) Branch
wise
Analysis.
-1
(seco , q is the
ferent
time
intervals
t
,
t
,
t
,
…
with
speeds
v
,
v
,
4
ms
for
5
minute
and
a
1
2
3
1
2
n
ngle
d) A
A
a
=
2
a
⇒
(b)
Law
of
Conservation
of
Length
of
strings
2
1
E
⇒
C
B
A
t
w0 is
raver
v3, … respectively, then the
of particle averof 1 ms for 3 minutes.
⎛
w =speed
the in
sed (
connecting
the
pulleys.
⎜
in
itial a
w0be- given
kt ⎞ 2 as
radia aged over the total distance⎝can
(d) For a moving pulley,
Addition of Vectors
w is t
ngula
n)one
(Please consider all unifo
he fi
The passing over
(c) If
end of a string
a moving
pulin tim
E
B
r vwe
nal aof vectors
body
2 d⎠⎟ + d + .......
B
and number
ley equals the avera
(b) To find the sum of any
elocit
eTotal
ngula
d
+
t then
distance
covered
speeds
in respective inte
w
ley
is
fixed,
the
acceleration
of
the
other
A+
1
2
3
y
(
ill=st
in ra -1 vav =
r veloline
the left and the right
must represent the vectors
a is t by the directed
ot p+w
ds end
c
05_Kinematics 2_Part 1.indd 15
11/7/2019 12:08:38 PM
h
it
Total
time
elapsed
+
+
t
t
......
(free
or
connected
to
something)
is
twice
)
e
y
h
1
2 en3
(b)terminal
angu of the previous ( in r
,
segments
with
the
point
B
w
k
I
SOLuTION
f a is
ads -1 the acceleration of the 0moving
lar a
R
- t pulley. Following
… x + x2
) at t
ccele then
no a next
xP (=1) 1 Total
vectors as the initialfrpoint
d1 + d2 + d3 + ...... 2 = 0
ratio
cons vector,
im
ee to oft the
D
situations
this to be made as a good rule for
n ( in
vave t= show
tanpoint
Average
Speed
= 2
se initial
t , an of the
O
A
⇒
the lines segment joininguthe
d
d
d
r
A
ads -2
1
alyse
To
w0
applications.
+ 2 + t =3 +2 ......
)
.
t
⇒
x
=
2
x
h
first vector to the terminal
e
1
P
prob
v1 v2 v3
w = dq
O A + AB + BC + CD + D E = OE
k
lem a
Now
DATUM
2v1v2
or
(cC) T
n
d
aver changing with time then
CASE-1: vav =
=
he SI d dt D a = w
el
(e) dIffespeed
is icontinuously
age
s
d
v
+
w
=
unit
angu
1 v2
dt w
l1
of q is
(d) w 2
l
xP
a
is given by the closing side or the nth side of the polyr
v
d
π
radia
q
eloci
=
Average Velocity =
vdt l2
n, w
ty ov
gon taken in opposite order. So,
is rad -1
eT = 2cπ f ,
+
+
2 w
er th
wher
vav =
s an
0
is tim
{
and
e
P
d
a
k
T
f
a is r
e int
is theF
2 w
is the
dt
R = A+B+C+D+E
ads -2
erva
0
(e) H
frequ B
perio
Ee
l
v1 + v2
.
k
ency
d of
re fto
w
2
dt
CASE-2: vav =
=
revo
and
o, t
w
fello
lutio
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 15
11/28/2019 7:53:54
f= 1
2
0
ws ing he graphb
(f)n When a particle=moves
with a speed⎜⎛ v1wfor half
1 PM
1
.
2
k the
PRoPERtiES oF VECtoR ADDition
(
)
(
)
B+
A+
A+B+C+D
C
∫
∫
∫
∫
∫
∫
TENSION IN A LIGhT STrING
N1
N2
N2
W
(f)
xvi
N1 N2
1
2
is called
normal force N . Note that although N
the
The
force
exerted
at
any
point
in
the
light
rope/
and W are equal and opposite, they do not constitute
string/wire/rod
is called
an action-reaction
pair. the tension at that point.
We may measure the tension at any point in the light
rope by cutting a suitable length from Nit and inserting
a spring scale, then the tension is the reading of the
scale.
The tension is same at all points in the rope
only if the rope is unaccelerated and assumed to be
Chapter
Insight
massless.
W is exposed to
Whenever a thread or a rope or a wire
(a)
(b)
some kind of force,
a tension ( T ) develops
in it.
(a) A man standing on floor is in equilibrium.
(b) The free body diagram of the man gives W = N.
ing on the body, i.e., net force and torque acting on
the body have to be zero.
Equilibrium
Translational Equilibrium
∑
• ∑F = 0
• ∑F = 0
• ∑F = 0
•
n a body, it changes the
motion of a body or bodies
two type of forces.
nal forces are those which
e system, only action acts
on of these forces are not
nal forces are those which
the system bodies, hence,
ion of these forces are in
sider a situation, shown in
d over box B, and a force
Here system includes two
he force F which is acting
ystem is an external force
T shown as
(a) Tension in any branch of string must be
y
(a) In the arrangementChapter
shown,
for translational
6: Newton’s
Laws of Motion 6.69
equilibrium to exist, we have
and opposite frictional force on the table (as shown in
(a) A block of mass m supported by a spring is in equilibrium.
tending
toblock
drag gives
it in the
of the hori(b) The free figure),
body diagram
of the
kx = direction
mg.
zontal force we exert. This frictional force is due to
the bonding of the molecules of the block and the
11/26/2019 12:04:37 PM
table at the places where the surfaces
are in very close
contact. If we focus on the table only, we see that friction tends to move the table in the same direction in
which force F tends to move the block. This observa06_Newtons Laws of Motion_Part 1.indd 46
tion leads us to two very important conclusions.
F
y
z
Conceptual Note(s)
mg
Note(a)that in figure the block, of course
exerts an equal
(b)
f
x
z
a pair.
(The pair is shown with two arrows facing each
other). The frictional forces acting between surfaces at rest
withblocks
respect
each
other
are
of static
When two
(M1to
and
M2 say)
connected
byforces
a
m
mg called
friction.
force
static friction will
(a) The
(b)actsof
string are
pulled,
then maximum
tension on M
towards
1
theonsame
as the
smallest
necessary
to start
M2 andbe
that
M2 acts
towards
M asforce
shown
in
(a) A ball of mass m suspended
from the 1ceiling with an
motion.
Once motion has started, the frictional forces
theinextensible
figure.
string in equilibrium.
acting
(b) The free
bodybetween
diagram ofthe
the surfaces
ball gives Tusually
= mg. decrease so that
a smaller Tforce
T is necessary to maintain uniform
M2
F kx
M
1 The forces acting between surfaces in relative
motion.
motion xaremcalled forces of kinetic friction.
(b) If a body A
k attached to an ideal light inextensible
k
string, is pulledCwith
o na cforce
e pFtand
u ait lis placed
N o ton
e (as )
horizontal frictionless surface, then
f
∑t = 0
• ∑t = 0
• ∑t = 0
• ∑t = 0
•
x
Some other examples of static equilibrium are shown
o n c e pfigures.
tual Note(s)
in theCfollowing
l Note(s)
Rotational Equilibrium
F=0
∑
∑
When
two
bodies are kept in contact then each body
F
=
0
gives
∑
exerts a contact force on the other. The magnitudes of
Fx = 0 andFORCe
Fy = aND
0
CONTaCT
FRICTION
x
F1the
coscontact
a + F3 cos
γ - F4 acting
- F2 coson
β =the
0 two bodies
…(1) are equal
forces
in magnitude and opposite in direction. So the conFy = 0 gives
tact forces obey Newton’s Third Law.
F1 sina + F2 sin β - F5 - F3 N
sin=γnormal
= 0 force
…(2)
∑
F2
F1
v
α
β
F4
R = contact force
y
f = friction
x
γ
F3
F5
The contact force
acting on a particular body is not
necessarily perpendicular to the contact surface. So,
contact force can be resolved into two components.
Conceptual Notes
The Conceptual
Notes, Remarks,
Words of Advice,
Misconception
Removals provide
warnings to the
students about
common errors
and help them
avoid falling
for conceptual
pitfalls�
(a) Perpendicular to the contact surface and
(b) Parallel to contact surface.
The perpendicular component is called the Normal
11/26/2019 12:06:19
PM
Contact Force or Normal Reaction (generally
written
as N ) and the parallel component is called Friction
(generally written as f ).
Therefore if R is contact force then
R=
f 2 + N2
sTaTIC aND KINeTIC FRICTION
Chapter 5: Kinematics II
5.45
(a) friction does not always oppose motion (because
The frictional force between
two surfaces
before the
Solved
ProblemS
it is friction only which is trying to move the
relative motion actually starts is called static fricChapter 2: Measurements and General Physics 2.39
table)
tional force or static friction, while the frictional
Problem 1
The rangeforce
of the projectile is given by
(b) friction always opposes relative motion between
between two surfaces in contact and in relative motion
2
along a vertical circle of radius
2
sin θ cos θ
u
surfaces in contact (because it is tryingAtoparticle
move is moving
is called kinetic frictional force-1or kinetic friction.
R=
= 20 3
…(2)
r = 20 m with a constant speed v =solVeD
31.4 ms ProbleMs
as
the table in the same direction in which the block
Static friction is a self-adjusting
force and it adjusts g
shown in figure. Straight line ABC is horizontal and
has a tendency to move).
in magnitude
and A
direction
Its magpasses throughboth
the centre
of the circle.
shell is automatically.
fired
From (1) and (2), we get
ProbleM 1
q = -2applied
nitude is always equal to external ⇒
effective
from point A at the instant when the particle is at C.
th
( 2n - 1 )2
force, tending
cause thecentury
relative motion and its direcP.A.M. IfDirac,
a great
oftothe
AB isPhysicist
20 3 m and
the 20
shell collide with So, p = 4tan
distance
, qθ==-2 , s = -1 , r = -3
ReasONs FOR FRICTION
is always
to athat
of external applied force. 3
found that
from the
basicopposite
constants
numBtion
, then
prove
the particle
at following
Hence,
So, when
a body
not
in motion
or iswe
inget
equi-θ , n = 1
ber having
dimensional formula
same
as thatisof
time
(a) Frictional forces arise due to molecular
interacFor
smallest
2
( 2nlibrium,
- 1)
then force of static friction is equal
4 to
-2 -1 -3 -1
can be constructed
θ = i.e.,
tions due to which a bonding between
thetan
molt = e ε 0-2 mthe
e mp c c G
π
3
Chapter End Solved
component
of the applied External force(s)
ecules of the two surfaces or objects
incharge
contact
⇒ θtangenmin =430° =
(a) thewhere
on
the
electron
e
3
n is antial
integer.
to theFurther,
surface.show that smallest
e
comes into being due to which(b)
it becomes
diffiProblems
⇒
t
=
…(7)
permittivity
of
free
space
ε
o
value of θ is 30°.
Applied Externalε 02 me2⎞mpGc 3
cult to move one surface on the (c)
other.
mass of the electron m⎛e Force of ⎞ ⎛
Problem 2
v
⎜
⎟
(b) Interon
locking of extended parts of
object
intoprotonu m
These are based
= Force Tangential to the
(d)one
mass
of the
P⎜
An object⎟ A is kept fixed at the point x = 3 m and
⎝ Static Friction ⎟⎠ ⎜
the extended parts of the other object.
Contact
(e) speed of light c
⎝ surfaces inAlso,
⎠ on
r = 20 m
verify
the dimensional
of
Chapterabove
3:correctness
Vectors
3.25
y =you
1.25can
m
P raised
a plank
the ground.
multiple concept
θ
(f) Universal Gravitational
constant G.
equation
(7) tby
substituting
thestarts
dimensional
= 0,
the plank
movingformulae
along the
At time
C
A
B
O
-2
usage in a single
+x respective
direction with
an acceleration
of the
quantities
on the right.1.5 ms . At the
20 this
3 m Dirac’s number. You are
Obtain the relation for
problem approach
so as to expose a
student’s brain to
the ultimate throttle
required to take the
JEE examination�
06_Newtons Laws of Motion_Part 2.indd 69
same instant a stone is projected from the origin
SolVED
given that the desired number is proportional
to mp-1 PRoBlEMS
with
a velocity the
u as
shown.
A stationary
on
Now
substituting
values
of all
constants person
in equaand me-2 . What is the significance of this number?
the
stone hitting the object durSolution
PRoBlEM
1
B this time t as
tionthe
(7),ground
we11/26/2019
get observes
the12:52:02
value
of
Assume
constant
of proportionality to be 1.
v PM
ing its downward2 motion at an angle of 45° to the
At of
thethe
time
of firing of the
the particle
was
18
The sum
magnitudes
of shell,
two forces
acting
atat C
v1 the x-y plane. Find
t = 3.36 × 10
s ≅ motions
1011 years
All the
are in
solution the shell collides with it at B , therefore the num- horizontal.
a point and
is 18 and the magnitude of their resultant is
u and the time after which the stone hits the object.
ber
of
revolutions
completed
by
the
particle
is
odd
Dirac
estimated
this
time
to
be
the
age
of universe.
2
According
to the statement
thethe
problem,
have
12.
If the resultant
is at 90° of
with
force ofwe
smaller
C
A
Take g = 10 ms
.
O
( 2n - 1 ) of forces?
q -2 are
magnitude,
p what
-1 rthe
s magnitudes
,
where
n
is
an
integer.
multiple
of
half,
i.e.,
t ∝ e ε 0 me mp c G
ProbleM 2
y
2
v3
Solution
t be theoftime
period of revolution
the particle,
The mean life of the neutral elementary particle pion
TakingLet
constant
proportionality
as 1 of
(given
in
A
then
is 2 × 10 -7 ns . The1.25
agemof the universe is about 4 × 109
question)
and
dimensions
on both
sides,
Let
F1 and
F2 be
the two forces
where
F1 we
< F2get
P
Solution
q
year. Find a time that is midway between
these
two
N
p2π r -1 2-×3 3.14
-2
-1
2 4× 20
TF=+( AT
M4) s ( M ) × …(1)
t ) ( M= L A T ) ( =
Here,
v1
times oninthe
logarithmic
1 F2 = 18v
Change
velocity
= Δv = scale.
vu f - vi
31.4
v
( LTbe )diagrammati(M L T )
The statement to the question can
Now, for half revolution.
solution
If T be the time of flight of the shell, then this time T(a)
x
3m
cally drawn-as
shown in Figure.
O 4 q - r - 2s p + 2q
⇒ T = M q - 3 - s L-3 q + r + 3 s⎛T2pn+ and vhalf
v3 =
v time
v between these two on the
1⎞ A
of the particle Let Ift be
the
way
1 = revolutions
equals the time of ⎜
⎝ 2 ⎠⎟
Solution
logarithmic
Then are equal but v3 v
as
their scale.
magnitudes
Applying Principle of Homogeneity,
we get
Δv
( 2n - 1 ) F2
directions
opposite.
For
stone
to
hit
the
object,
log
t
+ log
t A, at time t
S
× 4 = 2 ( 2n - 1 ) s
- q⇒
- 3 -Ts== 0
…(1)
log e t = e 1 e 2 12
2
⇒ Δv = v3 - vy 1 2= v - ( -v ) = 2v
-3 q + r + 3 s = 0
…(2)
A
For a projectile, the time of flight is given by
Δv = 12v directed south.
⇒ log e t = log e ( t1t2 ) = log e t1t2 P
p + 4 q - r - 2s = 1 F1
…(3)
2u sin θ
2 mrevolution
1.25
(b) For, quarter
T=
p + 2q = 0
…(4)
g
45°
Taking antilog
we get
both
sides,
So, the diagram clearly
shows that 12 is the resultant
u
Δv = v2 - v1 and
θ = 90°
So, T = t
Adding
(3) for
eliminating
, we we
get observe
of
F1 and(2)F2and
(taken
in same
order).rAlso
t = t1t2
11/28/2019 7:53:59 PM
North
θ
that 12
q +i.e.,
p +resultant
s2u
= sin
1 θ is perpendicular to F . So, from
…(5)
-1 r
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 16
-1 3
-2 s
COLUMN-I
11. Statement-1:
A physical quantity
cannot be called as a
(r) 10 ms -1
vector if its magnitude is zero.
(A)
Particle
moving in
Statement-2:
A
vector
has
both,
magnitude
and
(D) Change in speed (nearly) (s) 6 ms -1
circle
direction.
(in magnitude)
COLUMN-II
17. Statement-1: If dot product and cross product of A
and B are zero, it implies that either of the vectors A
(p) a may be
and B must bea null vector.
perpendicular to v
Statement-2: Null vector is a vector with zero
Particle
12. Statement-1: The sum of two(B)
vectors
can moving
be zero. in
magnitude.
(q) a may
be in the
straight
line
9. Trajectory of particle launched obliquely
fromThe
thevectors cancel each
Statement-2:
other,
when they
direction of v
18. Statement-1: The cross product of a vector with itself
2
are equal and opposite.
x
, where, x and y are in
ground is given as y = x is a null vector.
(C) Particle undergoing
80 13. Statement-1: Two vectors are said to be like vectors if (r) a may make same
Statement-2:
projectile motion
v cross product of two vectors results
acute
angle with The
metre. For this projectile motion match the following if
they
have
same
direction
but
different
magnitude.
in a vector quantity.
g = 10 ms -2 .
Statement-2: Vector quantities
not have
specific
(D) do
Particle
moving
into (s) a may be opposite
direction.
space
19.
Statement-1: The minimum number of non coplanar
velocity
COLUMN-I
COLUMN-II
vectors whose sum can be zero, is four.
14. Statement-1: The scalar product of two vectors can be
The
Chapter
1: Mathematical
Physics
1.29
zero.
v atresultant of two vectors of unequal
11. 2.54
A body
is projected
from
the groundStatement-2:
with
velocity
(A) Angle of projection
(p)
20 m
Advanced
JEE Physics
magnitude
can be zero.
Statement-2: If two vectors an
areangle
perpendicular
to each
θ . Then match
the following.
of projection
(B) Angle of velocity with
(q)
80 m
other,
their scalar product will be zero.
20. Statement-1: If A ⋅ B = B ⋅ C , then A may not always
horizontal after 4 s
COLUMN-ICorreCt ChoiCe
COLUMN-II
multiple
type
QueStionS
praCtiCe
exerCiSeS
be equal to C .
15. Statement-1: Multiplying any vector by a scalar is a
(C) Maximum height
(r)
45°
meaningful
operations. This
Statement-2:
dot product
two vectors
involves
(A)section
Change
in momentum
(p) Remains
contains
Multiple Correct
Choice TypeThe
Questions.
Each of
question
has four
choices (A), (B), (C) and (D), out of
Statement-2: Taking dot product
a scalar
and a is/are
vec- correct.unchanged
cosine of the angle between the two vectors.
which of
ONE
OR MORE
(D)CorreCt
Horizontal range
Single
ChoiCe type(s)
QueStionS
tor
is
meaningless.
-1 ⎛ 1 ⎞
tan ⎜ ⎟
1. (B)AAngle
student
discussing the
of a medium
(C) Energy and Young’s modulus
⎝ 2⎠
atwhen
the highest
(q)properties
Independent
of
This section contains Single Correct Choice Type Questions. Each question
has four
choices
(A), (B), (C)projected
and (D),velocity
out of
(except
vacuum)
writes
(D) Light year and wavelength
point
which ONLY ONE is correct.
linked CoMprehenSion typeVelocity
QueStionS
of of
light
of lightpoint
⎛
⎞ ⎛ Velocity
⎞
8. The pairs of physical quantities that possess same
(C) Kinetic
energy
body
(r) At highest
2
=
⎜ in vacuum
1. If ax + bx + c = 0 , a ≠ 0 , then This section contains Linked
10. log
10 100 =⎝
m
is/are of a
Comprehension
Type Questions
oris Paragraph
Each set consists
⎠⎟ ⎝⎜ in mediu
⎠⎟ based Questions.dimensions
zero
(A) only
Renold’s
and coefficient of friction
Paragraph followed
by questions.
choices
one number
is correct.
(A) Each
1This question
2 (A), (B), (C) and (D), out of which
formula ishas four (B)
b ± b 2 − 4 ac
b ± b 2 − 4 ac
(D)
Horizontal
component
(s)
Minimum
at
(B) options)
Curie and frequency of light wave
(A) x = −
(B) (For
x = −the sake of competitiveness
be a few questions
(C)there
3(A)may
(D) that
4 may have more than one correct
dimensionally
correct
2
4
of velocity
highest point
(C) Latent heat and gravitational potential
incorrect
11. log e ((B)
mn ) dimensionally
=
4. The length of the diagonal found
PROBLEM
2 is
(D) inPlanck
constant
torqueLaws of Motion 6.175
Chapter
6:and
Newton’s
Comprehension
b ± b 2 − 4 ac
b ± b 2 − 4 ac 1
(C) numerically incorrect
(C) x = −
(D) x = −
(A) log
log
m
×
n
e
e
(D) dimensionally
and numerically
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7 units
10 units of length are expressed as G x c y h z ,
2a
a
9. (B)If dimensions
Consider a4parallelogram
whose adjacent
sides are given
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where G , c and h are the universal gravitational
anSwer
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13 units
2. integer/numeriCal
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given
as
= At
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⎛m
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(B)
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tively, (A),
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5.
The
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log
In this section, the answer to each question
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obtained
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This section
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Reasoning
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tion, answer
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STATEMENT 2. You have1 to mark
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1 your answer as 1
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by
Chapter
and(B)
General
2.65
⎛ 1⎞
2
2
2
(D)
log
m
−
log
3. 1. log
x + log
y = from a gun from the
3.
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ise fired
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⎜
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e
e
(B)
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A
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Chapter
4:TRUE
Kinematics
I ˆ4.113
If both
30° and the
(C)
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sq. units
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angle of the barrel to the horizontal 60° . The
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In
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Rate
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bles
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vertical
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7
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5. bubbles
log
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1
Chapter
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Measurements
and
General
Physics
2.73
e
10
(D)
-1 of
4 2unit(s),
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kg
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of time
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bm
11.
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Chapter
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3.50
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05_Kinematics 2_Part
11/7/2019 12:12:57
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Then
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COLUMN-II.
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Which
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then
z
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z
=
=
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±
1
±
33.
[2010]
⎜
⎟
COLUMN-I
COLUMN-II
Statement-2:
Friction
force
is
needed
to
stop
the
car
or
13. (D)
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of mass
m the
is placed
on a smooth
8.
log a x equals
1
3 y ⎠⎟ as
1 table.frequency.
⎠velocity
plasma
-2(A)
None of the
three pairs
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The
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The
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To
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4
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4
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The
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log
x
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5 e6
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Let Pressure
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The
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The
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6. ⎛ 1Statement-1:
Two
teams
having
aneeds
tug oftowar
always
⎝ a⎠
1
(coulomb)
G
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11/11/2019 4:14:01 PM
04_Kinematics 1_Part 4.indd 113
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(C) 0.0202_Measurements, General Physics_Part
(D) 0.01
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02_Measurements, General Physics_Part 2.indd 68 ⎝ ε kBT ⎠
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measurement
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Chapter Insight
xvii
Practice Exercise
(
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Inclusion of all
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From this fully
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02_Measurements, General Physics_Part 2.indd 73
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 17
11/26/2019 5:08:18 PM
11/28/2019 7:54:06 PM
xviii Chapter Insight
Chapter 6: NewtoN’s Laws of MotioN
1.
(c)
Since Δp = F Δt
FO=
4.
Hints and H.301
Explanations
From the graph we observe that the peak force
exerted on the ball is 16000 N = 16 kN.
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=
Δt
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Exhaustive
Hints and
Explanations h.255
solutions
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⇒ Δpy = 0
TN
⇒ F = 15
v
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Further by definition intensity of a photon beam is
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Similarly, Δpx = m ( vx - ux )
ever
defined as the
energyneeded),
incident per unit area of a surforce of equalθ magnitude back on the hose, hence the
2
V
(
)
(
)
⇒ Δpx = mv [ - sin 30° - sin 30° ]
f
gardener
a 15 N force in the direction of F
f face per unitFtime.
MgsinaθAmust
=Mg0 apply
help
students
Mgcosθ
θ = 45°
4
b
the velocity of water stream to hold the hose in its
E
⇒ Δpx = -2mv sin ( 30° ) = -30 kgms -1
θ
⇒ I=
θ
Hence, (B) andposition
(C) are (stationary).
correct.
AΔtenhance their
Hence, the correct answer is (C).
Moving upwards
Just remains stationary
30
4. All accelerated
frames
Non-inertial
frames.
5. When
theare
diver
falls freely,
then Since
the velocity of the
E
So, Fav =
= 150 N
problem-solving
= IA
…(5)
0.2 Choice Type Questions
earth rotates
about
own
axis
around
cos
q
F1 = mg sin q⇒+ mmg
Multiple
Choice
Type
Questions
diver its
justCorrect
before
heand
hitsrevolves
the surface
of
water is
Single Correct
Δt
the sun, so it is a Non-Inertial frame (STRICTLY
skills�
1
2.
of the
ball just
before hitting the floor is
(
)(
)
1.
(A)
is
correct
when
the
bicycle
is
being
pedalled
Substituting
(5)
in
(4),
we
get
u
=
2
gh
=
2
9
.
8
10
=
14
ms
F
=
mg
sin
q
−
m
mg
cos
q
1. Velocity
Maximum
possible
error
2
SPEAKING), whereas for a good number of cases we
the forceasis acting in the
assume
the earthbecause
to be anduring
Inertialpedalling,
frame of reference
v1Δ=H 2 gh1 , downwards
v12 - 0 2Δ=
{∵× 100
2IA
ΔΙ
ΔR
t 2 gh
1}
Given
) rear wheel and thus
p
- uthe
m ( von
backward Δdirection
fric-that F1 = 3 F2 tan q =
× 100
= 2 × 100 +
+
× 100
F = ( 0.6 ) ( 25 - 0 )
{
∵
⇒
}
x = 2.5 ( −4 ) ( 0.37 − 1 )
m ⇒ x = 6.30 m
= 0.6
Δt
⇒ x=6m
1
s
Rounded off value to
ree Significant Figures
7.36
2
CHAPTER 6
θ
⇒
CHAPTER 6
Now, Δpy = m ( vy - uy ) = mv [ cos ( 30° ) - cos ( 30° ) ]
Now F = is extremely
value of acceleration
small.
=
mgc
Ι
t the
maxjust after
VelocityHof the ball
impactRwith the floor
is
tional
Δtacts in the
Δt forward direction.
Hence,
(B) and (D)
areforce
correct.
⇒ ( sin 45° + m cos 45° )
2
∵
H
=
Ι
Rt
R
t
Also from (1), we get
(D) is correct when bicycle is not pedalled.
2
2
v2 = 2 gh2 , upwards
∵
0
2
v
=
g
h
( 60 ) ( 4 - 14 )
( ) 2
2
9.44
⇒ Answer
⇒ F1 = 3 ( sin 45° − m cos245° )
F =(A) andType
Hence, the correct answer is (D).
Integer/Numerical
Problems
Hence,
1(D) are correct.
F = ( IA )
From Impulse Momentum theorem,
c
15.8
On solving, we get
kg
Change
Momentum
2. ⇒All
1. Linear impulse,
J = accelerated
mv
2. The Impulse
MKS unit= of
η = in is
kgm −1 s −1
F 0= 600 N frames are Non-inertial frames. Since
F 2I
ms
m = 0.5
earth rotates about its own axis and revolves around
7.37
⇒ Radiation Pressure = =
⇒ I = m ( v2 + v1 )
J
−1
A c
g
−1 −1
the
so it is a Non-Inertial frame (STRICTLY
= 2.Your
5 mssun,
⇒ v0 = Test
Concepts-II
is
gcm
s
The
CGS
of
η
=
⇒
N = 10 m = 5
m
9.44
150
cms
Hence, (A) and (C) are correct.
SPEAKING),
whereas for a good number of cases we
(Based
on Constraints)
⇒ I=
2 ( 10 )( 20 ) + 2 ( 10 )( 5 )
as and Reasoning Type Problems
1000 kg cms
⇒ v = v0 e − t τ assume the earth to be an Inertial frame of reference
15.8
Assertion
ηMKS
4. When P is gradually increased, so till the moment P
1. Asthe
already
done
in an Illustration,
we obtained
=
×
value
of
acceleration
is
extremely
small.
equal
f0, mass A will
η 150 ms
g
dx
1. Both Statementsbecomes
are correct.
Butto
Statement-II,
doesnot
notmove and hence
⇒ I =CGS ( 20 + 10 ) = 4.5 kgms -1
aB + (B)
aC +and
2 aA(D)
= 0 are correct.
⇒
= v0 e − t τHence,
use the least number of signifithe
string will not develop any tension due to pull P.
1000 3
explain
correctly,
Statement-I.
dt
10
g
×
cm
×
s
η
MKS
e decimal is 1.
P <anf0increase
. Furtherinfor
A and B to move colSo, T =There
0 for is
thecan
force
exerted on the mirror by the photon.
3. Similarly,
Let F bewe
=
find
-1 = 10
Correct explanation:
normal
So, impulse
4.5cm
kgms
, upwards
×s×g
ηCGS is100
τ
x
with
(say), is
weamust have
Since a photon will be reflected from the mirror
with when lectively
the least number of significant
reaction
the object
is acceleration
pushed anda there
− t τ v + v + 2v = 0
⇒
dx = v0 ethe dt
B
CvalueA of momentum so, change in momen
same
momencimal is 1.
Hence,
theimpulse
correct answer
3. (a)
Since
is equalisto(A).
the area under the F-t
decrease in normal reaction
is pulled
P - (when
f1 )lim the
- ( fobject
2 )lim
2E
0
0Taking, upward direction as
a=
we are given:
se the least number of signifigraph
(but strictly not horizontally).
tum Δ
Δp
p = p - ( - p ) = 2 p = positive
.
3.
i=K
tan θ
2m
c
− t τ ⎤τ v = v = 1 ms -1
Since tan θ is a dimensionless physical quantity and
⎡
e
2.
The
cloth
can
be
pulled
dislodging the
A
B
Pout
- 2 fwithout
I = Fdt = Area under F-t graph
0
⇒
x
=
v
use the least number of signifi⎢
⎥
⇒
a
=
…(1)
0
hence the unit of k is equivalent to that of current i.e.
(Because for a photon E = pc )
dishes from the table due 2to
m law of inertia, which is
⎣ − 1⇒τ ⎦ 0 v = -3 ms -1
not 2), because 1100 ms −1 has
ampere
C
Newton’s
First
Law.
While,
the
Statement-II
is
true,
1
So, the system will not move till P > 2 f0 , but T will
( 16000 )
⇒ theI correct
= ( 3.5answer
- 1 ) ( 10 -is3 )(B).
ures.
Hence,
but it is Newton’s Third Law.
) ( e −1 − e 0 )
⇒ x = 2.5 ( −4i.e.,
2
velocity of block C is 3 ms -1 (downwards).
be developed as soon as P becomes greater than f0 .
T
4.
Impulse = change in momentum
θ
Hence
Tcosθ
⇒ I = 20 Ns
T
{towards right}
2.
vA = 2 ms -1
division on main scale
So, the dimensions of impulse and momentum are the
P = T + f0
(b)
Since we know that Fav Δt = ΔI
same.
divisions on vernier scale
Tsinθ
vA
-1
⇒
{upwards}
v
=
=
1
ms
Hence, the correct answer
is
(C).
⇒
T
= P - f0 for f0 < P < 2 f0
F
P
1
2 Light
⇒
Fav ( 2.5 × 10 -3 ) = 20
m = 0.1 mm
5. Coefficient of viscosity
So, (B) is also correct.
-1
mg
vB = 2 ms
{towards left}
⇒
Fav = 8000 N = 8 kN
m
tangential force
Once P > 2 f0 , then for
η=
Δ
p
contact area × velocity gradient
⇒ F=
BLOCK A
.C. ) = ( 10.2 + 3 × 0.01 ) cm
Δt
⇒ η = newton = newton s
P - T - f0 = ma
…(2)
s
2E
m2
m2 × m
…(1)
⇒ F=
P - 2 f0 ⎞
⎛
m
cΔT
⇒ P - T - f0 = m ⎜
{∵ of ( 1 ) }
⎝ 2m ⎟⎠
kg ms kg
Also
η= 2 2 =
P
ms
s m
6 (pitch) = 6 mm
⇒ P - T - f0 = - f0
T cos q = mg and
…(2)
06_Newtons Laws of Motion_Solution_P1.indd 193
11/26/2019 12:35:26 PM
2
Hence, the correct answer is (B).
= n ( L.C. ) = 40 × 0.01 = 0.4 mm
2E
P
⇒
T
sin
q
=
F
=
…(3)
P
T
=
0.4 ) = 6.4 mm
an2
cΔT
2
6. Pressure correction P ′ = 2
V
ale lies to the right of zero of
P
2E
⇒ T=
…(4)
⇒
tan
q
=
s positive zero error.
2
2
(
)(
)2
mgcΔT
⇒ a = P ′V = Pressure correction Volume
Hence, (A), (B) and (C) are correct.
2
n2
gas ) Laws of Motion_Solution_P2.indd 301
( amount of06_Newtons
er constant is 0.01 cm
11/26/2019 12:42:56 PM
8.33
{
{
(
}}
CHAPTER 6
Test Your Concepts-I
(Based on Impulse Momentum)
)
∫
∫
∫
+ x × V.C.
⇒
× 0.01
m
.05 cm
h will be 0.05 cm less than the
2
⇒
a=
a=
( Nm −2 ) ( m 3 )2
mol 2
kgmm 4
Nm
=
mol 2 s 2mol 2
4
2
{∵ 1 N = 1 kgms−06_Newtons
} Laws of Motion_Solution_P1.indd 255
11/26/2019 12:44:31 PM
=Ι
11/11/2019 4:37:26 PM
p= p-
{
=2 =
}
)
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 18
11/28/2019 7:54:11 PM
Preface
In the past few years, the IIT-JEE has evolved as an examination designed to check a candidate’s true
scientific skills. The examination pattern needs one to see those little details which others fail to see. These
details tell us how much in-depth we should know to explain a concept in the right direction. Keeping the
present-day scenario in mind, this series is written for students, to allow them not only to learn the tools but
also to see why they work so nicely in explaining the beauty of ideas behind the subject. The central goal of this
series is to help the students develop a thorough understanding of Physics as a subject. This series stresses on
building a rock-solid technical knowledge based on firm foundation of the fundamental principles followed by
a large collection of formulae. The primary philosophy of this book is to guide the aspirants towards detailed
groundwork for strong conceptual understanding and development of problem-solving skills like mature and
experienced physicists.
This updated Third Edition of the book will help the aspirants prepare for both Advanced and Main levels
of JEE conducted for IITs and other elite engineering institutions in India. This book will also be equally useful
for the students preparing for Physics Olympiads.
This book is enriched with detailed exhaustive theory that introduces the concepts of Physics in a clear,
concise, thorough and easy-to-understand language. A large collection of relevant problems is provided in
eight major categories (including updated archive for JEE Advanced and JEE Main), for which the solutions are
demonstrated in a logical and stepwise manner.
We have carefully divided the series into seven parts to make the learning of different topics seamless
for the students. These parts are
•
•
•
•
•
•
•
Mechanics – I
Mechanics – II
Waves and Thermodynamics
Electrostatics and Current Electricity
Magnetic Effects of Current and Electromagnetic Induction
Optics
Modern Physics
Finally, I would like to thank all my teacher friends who had been a guiding source of light throughout the
entire journey of writing this book.
To conclude, I apologise in advance for the errors (if any) that may have inadvertently crept in the text.
I would be grateful to the readers who bring errors of any kind to my attention. I truly welcome all comments,
critiques and suggestions. I hope this book will nourish you with the concepts involved so that you get a great
rank at JEE.
PRAYING TO GOD FOR YOUR SUCCESS AT JEE. GOD BLESS YOU!
Rahul Sardana
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 19
11/28/2019 7:54:11 PM
About the Author
Rahul Sardana, is a Physics instructor and mentor having rich and vast experience of about 20 years in the
field of teaching Physics to JEE Advanced, JEE Main and NEET aspirants. Along with teaching, authoring
books for engineering and medical aspirants has been his passion. He authored his first book ‘MCQs in
Physics’ in 2002 and since then he has authored many books exclusively for JEE Advanced, JEE Main and
NEET examinations.
He is also a motivational speaker having skills to motivate students and ignite the spark in them for
achieving success in all colours of life. Throughout this journey, by the Grace of God, under his guidance and
mentorship, many of his students have become successful engineers and doctors.
F01_Physics for JEE Mains and Advanced_Mechanics I_Prelims.indd 20
11/28/2019 7:54:11 PM
CHAPTER
1
Mathematical Physics
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Quadratic Equations
(f) Geometric Progression (GP) (k) Functions
(b) Logarithmic Functions
(g) Coordinate Geometry
(l) Limit of a Function
(c) Linear Equations
(h) Trigonometry
(m) Differentiation
(d) Determinants
(i) Factorial
(n) Integration
(e) Arithmetic Progression (AP) ( j) Series Expansions
All this is followed by an Exercise Set which contains questions for your practice. Please be advised
that you must first thoroughly study this chapter for your command on the Mathematical tools
used in Physics hereafter.
GEnERAL MAtHEMAtICs: A REVIEW
This part in mathematics is introduced here to give
you a fundamental review of operations and methods.
To have an extra understanding in Physics you should
be totally familiar with basic algebraic techniques,
analytic geometry, and trigonometry. Differential
and integral calculus are discussed in detail and are
intended for those students who have difficulties in
applying calculus concepts to physical situations.
Table 1.1
Mathematical symbols used in the text and their
meaning
Symbol
Meaning
Table 1.1 (Continued)
Symbol
Meaning
()
much greater (less) than
≈
is approximately equal to
∼
positive difference between two numbers
Δx
the change in x
N
∑x
i
the sum of all quantities xi from i = 1 to
i=N
i =1
x
the magnitude of x (always a positive
quantity)
=
is equal to
Δx → 0
Dx approaches zero
≠
is not equal to
∝
the derivative of x with respect to t
is proportional to
dx
dt
>
is greater than
∂x
∂t
the partial derivative of x with respect to t
<
is less than
∫
Integral
(Continued)
01_Mathematical Physics_Part 1.indd 1
11/28/2019 6:43:58 PM
1.2 JEE Advanced Physics: Mechanics – I
ALGEBRA
⇒
When algebraic operations are performed, the laws
of arithmetic apply. Symbols such as x , y and z
are usually used to represent quantities that are not
specified, and symbols such as a , b and c are used
to represent numbers.
You should be familiar with the following operations:
x1 =
27 + 23 5
27 − 23 1
= and x2 =
=
20
2
20
5
Roots of the equation are
MULTIPLYING POWERS OF A GIVEN
QUANTITY
Fractions
(i) x n x m = x n + m
⎛ b ⎞ ab
(i) a ⎜ ⎟ =
⎝ c⎠
c
⎛ a ⎞ ⎛ c ⎞ ac
(ii) ⎜ ⎟ ⎜ ⎟ =
⎝ b ⎠ ⎝ d ⎠ bd
⎛ a⎞
⎜⎝ ⎟⎠ ad
b
=
(iii)
⎛ c ⎞ bc
⎜⎝ ⎟⎠
d
(iv)
(iii) ln(e ) = 1
(iv) ln(e x ) = x
(v) ln(xy ) = ln x + ln y
Roots of a Quadratic Equation
⎛ 1⎞
(vii) ln ⎜ ⎟ = − ln x
⎝ x⎠
−b ± b − 4 ac
x=
2a
If α and β are two roots of equation, then
α +β = −
b
c
and αβ =
a
a
Illustration 1
Solve the equation 10 x 2 − 27 x + 5 = 0 .
Solution
By comparing the given equation with standard
equation a = 10 , b = −27, and c = 5.
x=
⇒
−b ± b 2 − 4 ac
2a
−( −27 ) ± ( −27 )2 − 4 × 10 × 5 27 ± 23
x=
=
2 × 10
20 01_Mathematical Physics_Part 1.indd 2
= xn− m
(ii) log = logarithm to base 10
⎛ x⎞
(vi) ln ⎜ ⎟ = ln x − ln y
⎝ y⎠
2
x
m
(i) ln = logarithm to base e , also called as natural
log.
(iii) x 2 − 2x − 15 = (x + 3)(x − 5) {Quadratic Equation}
If ax 2 + bx + c = 0 ( a ≠ 0 ) , has two roots given by
xn
Logarithmic Functions
Factoring and Combinations
(ii) x 2 − y 2 = (x + y )(x − y )
(ii)
(iii) (x n )m = x nm
a c ad ± cb
± =
b d
bd
(i) ax + bx = x( a + b)
5
1
and .
2
5
( )
(viii) ln x n = n ( ln x )
(ix) ln ( a ) = 2.3026 log ( a )
(x) ln a = log e a = 2.303 log10 a
(xi) log a x =
log e x
log e a
(xii) If log e x = α , then x = eα
Simultaneous Linear Equations
In order to solve two simultaneous equations involving two unknowns, x and y , we solve one of the
equations for x in terms of y and substitute this
expression into the other equation.
Illustration 2
Solve the following given equations
5x + y = −8 …(1)
2x − 2 y = 4 …(2)
11/28/2019 6:44:07 PM
Chapter 1: Mathematical Physics
If numbers written as powers of ten are multiplied,
we simply add the exponents, maintaining their
signs. For example,
Solution
From (2), x = y + 2, substituting this in (1) gives
5 ( y + 2 ) + y = −8
⇒
⇒
⇒
6 y = −18 (3 × 10 3 ) × (5 × 10 4 ) = 15 × 107 = 1.5 × 108
(2 × 10 5 ) × ( 4 × 10 −2 ) = 8 × 10 3
y = −3 (5.6 × 10 4 ) × ( 4.3 × 108 ) = 24 × 1012
x = y + 2 = −1 When numbers written as powers of ten are divided,
we can bring the power of ten from the denominator
to the numerator by changing its sign. For example,
Alternate Solution:
Multiplying (1) by 2 and add the result to (2)
10 x + 2 y = −16
2x − 2 y = 4
⇒
⇒
8 × 10 5
2 × 10 2
12x = −12 4 × 10 −9
In general,
2nd order (four elements) i.e., a 2 × 2 determinant is
evaluated as
a2
= a1b2 − a2 b1
b2
3 order (nine elements) i.e., a 3 × 3 determinant is
evaluated as
a3
b2
b3 = a1
c2
c3
10n
10 m
= 10n − m
(10n )m = 10nm
ARITHMETIC PROGRESSION (AP)
rd
a2
b2
c2
= 3 × 10 −4 × 109 = 3 × 10 5
10n10 m = 10n + m
Determinants
a1
b1
c1
= 4 × 10 5 × 10 −2 = 4 × 10 3
12 × 10 −4
x = −1 y = x − 2 = −3 a1
b1
b3
b1
− a2
c3
c1
b1
b3
+ a3
c1
c3
b2
c2
= a1 ( b2 c3 − b3 c2 ) − a2 ( b1c3 − b3 c1 ) + a3 ( b1c2 − b2 c1 )
It is a sequence of numbers which are arranged in
increasing order and having a constant difference
between them.
EXAMPLE: 1, 3, 5, 7, 9, 11, 13, … or 2, 4, 6, 8, 10, 12, …
In general arithmetic progression can be written as
a0 , a1 , a2 , a3 , a4 , a5 …
(a) nth term of an arithmetic progression is given by
an = a0 + ( n − 1 ) d
POWERS OF TEN
You should be familiar with the usage of powers of
ten. It is a compact form of writing very large or very
small numbers. For example, instead of 10000, we
write 10 4, where the exponent represents the number of zeros; that is 10 4 = 10 × 10 × 10 × 10 = 10000.
Likewise, a small number like 0.0001 can be expressed
as 10 −4, where the negative exponent indicates that
we are dealing with a number less than one. Some
other examples of the use of powers of ten are
where a0 is the first term, n is the number of
terms, d is the common difference given by
d = ( a1 − a0 ) = ( a2 − a1 ) = ( a3 − a2 )
(b) Sum of arithmetic progression
n
n
Sn = ( 2 a0 + ( n − 1 ) d ) = ( a0 + an )
2
2
Illustration 3
Find the sum of series 7 + 10 + 13 + 16 + 19 + 22 + 25 .
1000 = 10 3
0.003 = 3 × 10 −3
Solution
85 000 = 8.5 × 10 4
0.00085 = 8.5 × 10 −4
Since n = 7 , a0 = 7 and an = a7 = 25, so
3 200 000 = 3.2 × 106
0.00002 = 2 × 10 −5
01_Mathematical Physics_Part 1.indd 3
1.3
Sn =
7
n
( a0 + an ) = ( 7 + 25 ) = 112
2
2
11/28/2019 6:44:13 PM
1.4 JEE Advanced Physics: Mechanics – I
GEOMETRIC PROGRESSION (GP)
It is a sequence of numbers in which every successive
term is obtained by multiplying the previous term by
a constant quantity. This constant quantity is called
the common ratio ( r ) .
(b) Radian: The arc length l is proportional to the
radius r for a fixed value of θ (in radians)
l = rθ
⇒
θ=
l
r
EXAMPLE: 4, 8, 16, 32, 64, 128 … where a = 4 and r = 2
1,
In general geometric progression can be written as a ,
ar , ar 2 , ar 3 , ar 4 , … where a is the first term and r
is the common ratio
(a) Sum of n terms of G.P. is
if
r<1
Sn =
a (r n − 1)
r −1
if
r>1
(b) Sum of infinite terms of G.P.
S∞ =
a
1− r
if
θ
r
(c) Circumference of a circle is C = 2π r
(d) Area of a circle is A = π r 2
a(1 − r n )
Sn =
1− r
l
1 1 1 1
1
, , ,
where a = 1 and r =
2 4 8 16
2
1 2
r θ , where θ is in radian
2
(f) The pythagorean theorem, which relates the
three sides of a right triangle
(e) Sector area =
c2 = a2 + b 2
r<1
c
b
Illustration 4
Find the sum of series Q = 2q +
q q q
+ +
+ ......
3 9 27
Solution
Above equation can be re-written as
q q q
⎤
⎡
Q = q+ ⎢q+ + +
+ ...... ⎥
3 9 27
⎣
⎦
By using the formula of sum of infinite terms of G.P.
3
5
⎡ q ⎤
Q = q+ ⎢
= q+ q = q
1⎥
2
2
⎢ 1− ⎥
3⎦
⎣
COordinate GEOMETRY
(a) The distance d between two points whose coordinates are ( x1 , y1 ) and ( x2 , y 2 )
d = (x2 − x1 )2 + ( y 2 − y1 )2
01_Mathematical Physics_Part 1.indd 4
a
(g) Area of a triangle is A =
1
( Base )( Altitude )
2
(h) Surface area of a sphere is A = 4π r 2
4 3
π r . If a spherical
3
shell (hollow sphere) of radius x and thickness
dx is cut out from the centre then the surface
area and volume of the solid part is given by
4π r 2 and 4π x 2 dx respectively
(i) Volume of a sphere is V =
(j) Volume of a cylinder is V = π r 2 l
(k) Equation of a straight line is y = mx + b ( b = y
intercept, m = slope = tan θ ). The equation of a
x y
straight line in intercept form is + = 1 .
a b
11/28/2019 6:44:20 PM
Chapter 1: Mathematical Physics
y
y
m
θ
b
(o) Equation of a
xy = constant .
pe
lo
=s
rectangular
hyperbola
1.5
is
y
b
O
x
a
x
O
(l) Equation of a circle of radius R centred at the origin is x 2 + y 2 = R2 . Equation of a circle of radius
R centred at ( a, b ) is the general form of equa-
Rectangular hyperbola
tion of a circle is
x 2 + y 2 + 2 gx + 2 fy + c = 0
This circle has centre at ( − g , − f ) with radius
r=
g2 + f 2 − c.
x
O
(p) For a right circular cone of height h, radius r, the
1
volume is V = π r 2 h , Lateral surface area = π rL,
3
where L = r 2 + h 2 .
( x − a ) 2 + ( y − b )2 = R 2
L
y
R
r
Right circular cone
b
O
(q) Every triangle inscribed within a semicircle is a
right triangle.
x
a
(m) Equation of an ellipse with the origin at its ­center
x2 y2
is 2 + 2 = 1 ( a = semi-major axis, b = semia
b
minor axis).
y
b
0
h
θ1
θ2
θ 1 + θ 2 = π /2
Trigonometry
a
x
(a) The angle θ is measured in radian.
180° = π radian
Ellipse
(n) Equation of a parabola whose vertex is at y = b is
y = ax 2 + b.
y
π
radian ≈ 0.02 radian
180
180
Similarly 1 radian =
degree ≈ 57°
π
⇒
1° =
π
180
180
Radian to Degree: Multiply by
π
Degree to Radian: Multiply by
b
O
Parabola
01_Mathematical Physics_Part 1.indd 5
x
11/28/2019 6:44:25 PM
1.6 JEE Advanced Physics: Mechanics – I
(b) Basic Trigonometric Quantities
The sine, cosine and tangent functions in
­trigonometry are defined in terms of the ratio of
the sides of a right triangle:
y
sinq
cosq
tanq
cosecq
secq
cotq
Quad III
(tan)
|
|
⊕
|
|
⊕
Quad IV
(cos)
|
⊕
|
|
⊕
|
P(x, y)
r
θ
O
(i) sin θ =
a Perpendicular ( P )
=
c
Hypotenuse ( H )
cosec 2θ = 1 + cot 2 θ
(iii) cot θ =
a Perpendicular ( P )
=
b
Base ( B )
(c) Signs in the four quadrants.
(+)
(+)
θ
1
tan θ
(f) The relations at the right follows directly from
the right triangle above:
In First Quadrant
sin θ = cos(90° − θ )
cos θ = sin(90° − θ )
cot θ = tan(90° − θ )
In Second Quadrant:
b
Base ( B )
=
c Hypotenuse ( H )
I
sin 2 θ + cos 2 θ = 1; sec 2 θ = 1 + tan 2 θ ;
sin θ
cos
θ
(e) The cosecant, secant, and cotangent functions are
defined by
1
1
(i) cosec θ =
(ii) sec θ =
sin θ
cos θ
Side opposite θ
Side adjacent θ
⇒ sin θ =
tan θ =
Side adjacent θ
Hypotenuse
⇒ cos θ =
(iii) tan θ =
x
x
Side opposite θ
Hypotenuse
⇒ sin θ =
(ii) cos θ =
(d) From the above basic units and using the
Pythagorean theorem, it follows that
y
sin ( 90° + θ ) = cos θ
sin ( 180° − θ ) = sin θ
cos ( 90° + θ ) = − sin θ cos ( 180° − θ ) = − cos θ
II
θ
(+) tan ( 90° + θ ) = − cot θ tan ( 180° − θ ) = − tan θ
In Third Quadrant:
(–)
sin ( 180° + θ ) = − sin θ
θ
(–)
(+)
θ
III
cos ( 180° + θ ) = − cos θ cos ( 270° − θ ) = − sin θ
IV
(–)
(–)
sin ( 270° − θ ) = − cos θ
sinq
cosq
tanq
cosecq
secq
cotq
Quad I
(All)
⊕
⊕
⊕
⊕
⊕
⊕
Quad II
(sine)
⊕
tan ( 180° + θ ) = tan θ
In Fourth Quadrant:
tan ( 270° − θ ) = cot θ
sin ( 270° + θ ) = − cos θ
sin ( 360° − θ ) = − sin θ
cos ( 270° + θ ) = sin θ
cos ( 360° − θ ) = cos θ
tan ( 270° + θ ) = − cot θ tan ( 360° − θ ) = − tan θ
(g) Some properties of trigonometric functions:
(i) sin( −θ ) = − sin θ
01_Mathematical Physics_Part 1.indd 6
|
|
⊕
|
|
(ii) cos( −θ ) = cos θ
(iii) tan( −θ ) = − tan θ
11/28/2019 6:44:28 PM
Chapter 1: Mathematical Physics
(h) Values for certain angles
MEASUREMENT OF POSITIVE and
NEGATIVE ANGLES
q (in
degree)
q (in
radian)
sinq
cosq
tanq
0°
0
0
1
0
30°
π
6
1
2
3
2
1
3
45°
π
4
1
2
1
2
1
60°
π
3
3
2
1
2
3
90°
π
2
1
0
Not
Defined
120°
2π
3
3
2
1
−
2
− 3
135°
3π
4
1
2
−
1
2
−1
150°
5π
6
1
2
3
−
2
1
−
3
180°
p
0
−1
0
3
Also take a note that sin ( 37° ) = cos ( 53° ) =
5
Sides and Angles of Triangle
(i) Sum of all the angles of Triangle is 180°.
Angles measured counterclockwise (CCW) from
the positive x-axis are assigned ­
positive measures whereas angles measured clockwise (CW) are
assigned negative measures.
y
y
x
Negative
measure
Positive
measure
x
(Counter Clockwise Sense) y
x
9π
4
3π
x
y
y
–
–
3π
4
5π
2
x
x
Some Trigonometric Identities
γ
a
b
β
α
c
(j) Law of cosine
(i) a 2 = b 2 + c 2 − 2bc cos α
(ii) b 2 = a 2 + c 2 − 2 ac cos β
(iii) c 2 = a 2 + b 2 − 2 ab cos γ
(k) Law of sines (Lami’s Theorem), for any triangle
a
b
c
=
=
sin α sin β sin γ
01_Mathematical Physics_Part 1.indd 7
(Clockwise Sense)
y
⇒ α + β + γ = 180°
1.7
(a) sin 2 θ + cos 2 θ = 1
(b) cosec 2 θ = 1 + cot 2 θ
(c) sec 2 θ = 1 + tan 2 θ
⎛θ⎞ 1
(d) sin 2 ⎜ ⎟ = (1 − cos θ )
⎝ 2⎠ 2
(e) sin ( 2θ ) = 2 sin θ cos θ
⎛θ⎞ 1
(f) cos 2 ⎜ ⎟ = (1 + cos θ )
⎝ 2⎠ 2
(g) cos ( 2θ ) = cos 2 θ − sin 2 θ
⎛θ⎞
(h) 1 − cos θ = 2 sin 2 ⎜ ⎟
⎝ 2⎠
11/28/2019 6:44:34 PM
1.8 JEE Advanced Physics: Mechanics – I
(i) tan ( 2θ ) =
2 tan θ
⎛ π ⎞
radian
(d) 135° = 135 × ( 1° ) = 135 × ⎜
⎝ 180 ⎟⎠
1 − tan 2 θ
1 − cos θ
⎛θ⎞
(j) tan ⎜ ⎟ =
⎝ 2⎠
1 + cos θ
⇒ 135° =
(k) sin( A ± B) = sin A cos B ± cos A sin B
(l) cos( A ± B) = cos A cos B ∓ sin A sin B
⎛ C+D⎞
⎛ C−D⎞
(m) sin C + sin D = 2 sin ⎜
cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎤
⎤
⎡1
⎡1
(n) cos A + cos B = 2 cos ⎢ ( A + B) ⎥ cos ⎢ ( A − B) ⎥
⎣2
⎦
⎣2
⎦
(o) sin ( 3θ ) = 3 sin θ − 4 sin 3 θ
⎛ π ⎞
radian
(e) 210° = 210 × ( 1° ) = 210 × ⎜
⎝ 180 ⎟⎠
⇒ 210° =
3 tan θ − tan 3 θ
⎛ π ⎞
(f) 225° = 225 × ( 1° ) = 225 × ⎜
radian
⎝ 180 ⎟⎠
⇒ 225° =
⇒ 270° =
1 − 3 tan 2 θ
(r) cos ( 2θ ) = cos 2 θ − sin 2 θ =
1 − tan 2 θ
2
9π
radian
4
1 + tan θ
2
(t) cos 2 A − sin 2 B = cos ( A + B ) cos ( A − B )
⎛ C+D⎞
⎛ C−D⎞
cos ⎜
(u) cos C + cos D = 2 cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎛ C−D⎞
⎛ C+D⎞
cos ⎜
(v) sin C − sin D = 2 sin ⎜
⎟
⎝ 2 ⎠
⎝ 2 ⎟⎠
⎛ C−D⎞
⎛ C+D⎞
cos ⎜
(w) cos C − cos D = −2 sin ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
3π
radian
2
⎛ π ⎞
(h) 300° = 300 × ( 1° ) = 300 × ⎜
radian
⎝ 180 ⎟⎠
(s) sin A − sin B = sin ( A + B ) sin ( A − B )
2
7π
radian
6
⎛ π ⎞
(g) 270° = 270 × ( 1° ) = 270 × ⎜
radian
⎝ 180 ⎟⎠
(p) cos ( 3θ ) = 4 cos 3 θ − 3 cos θ
(q) tan ( 3θ ) =
3π
radian
4
⇒ 300° =
5π
radian
3
⎛ π ⎞
radian
(i) 330° = 330 × ( 1° ) = 330 × ⎜
⎝ 180 ⎟⎠
⇒ 330° =
11π
radian
6
Illustration 6
Find the six trigonometric ratios from the given figure.
Illustration 5
5
Convert the following angles to radian.
(a) 45°
(d) 135°
(g) 270°
(b) 60°
(e) 210°
(h) 300°
(c) 120°
(f) 225°
(i) 330°
π
⎛ π ⎞
radian = radian
(a) 45° = 45 × ( 1° ) = 45 × ⎜
⎝ 180 ⎟⎠
4
π
⎛ π ⎞
(b) 60° = 60 × ( 1° ) = 60 × ⎜
radian = radian
⎝ 180 ⎟⎠
3
⎛ π ⎞
(c) 120° = 120 × ( 1° ) = 120 × ⎜
radian
⎝ 180 ⎟⎠
01_Mathematical Physics_Part 1.indd 8
12
Solution
By Pythagoras Theorem, we have
Solution
⇒ 120° =
θ
2π
radian
3
H 2 = P 2 + B2
⇒
⇒
H 2 = 52 + 122 = 169 H = 13 5
P
B 12
P 5
=
, cos θ =
=
, tan θ = =
H 13
H 13
B 12 ⇒
sin θ =
⇒
cosec θ =
13
13
12
, sec θ =
, cot θ =
5
12
5 11/28/2019 6:44:41 PM
Chapter 1: Mathematical Physics
Illustration 7
n
(b) ( 1 + x ) = 1 + nx +
Find the values of
(a) cos ( 120° )
(b) sin( −1485°)
(d) cos ( −60° )
(e) tan ( 210° )
n ( n − 1)( n − 2 ) 3
x + .....
3!
{Binomial Series}
(a) cos ( 120° ) = cos ( 90° + 30° ) = − sin ( 30° ) = −
1
2
(b) sin ( −1485° ) = − sin ( 3 × 360° + 45° )
⇒ sin ( −1485° ) = − sin ( 45° ) = −
1
2
− 3
(c) sin ( 300° ) = sin ( 360° − 60° ) = − sin ( 60° ) =
2
1
(d) cos( −60°) = cos ( 60° ) =
2
1
(e) tan ( 210° ) = tan ( 180° + 30° ) = tan ( 30° ) =
3
Illustration 8
If A = 60°, then find the value of sin ( 2A ) .
Solution
Since sin ( 2 A ) = 2 sin A cos A
x2 x3
(c) e x = 1 + x +
+
+
2! 3!
(e) sin x = x −
x3 x3
+
−
3! 5!
(f) cos x = 1 −
x2 x4
+
− { x in radian}
2! 4!
π
x 3 2x 3
+
+ x <
3
15
2
{Trigonometric Expansion}
(h) For x 1 , the following approximations can be
used:
(g) tan x = x +
(i) (1 + x )n ≈ 1 + nx {Binomial Approximation}
(ii) ln(1 ± x ) ≈ ± x {Logarithmic Approximation}
sin ( 2 A ) = 2 sin A cos A = 2 sin ( 60° ) cos ( 60° )
⎛ 3 ⎞ ⎛ 1⎞
3
sin ( 2 A ) = 2 ⎜
=
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
2 FACTORIAL
Factorial is always defined for a positive integral
number, say n . Then
n = n ! = n ( n − 1 ) ( n − 2 ) ( n − 3 ) .......... × 3 × 2 × 1
read as n
factorial
)
Also, we observe that n = n n − 1 = n ( n − 1 ) !
n
(a) ( a + b ) = an +
01_Mathematical Physics_Part 1.indd 9
n n −1 n ( n − 1 ) n − 2 2
a b+
a b + ⋅⋅
1!
2!
{Binomial Expansion}
(iii) e x ≈ 1 + x (iv) sin x ≈ x (v) cos x ≈ 1 {Exponential Approximation}
{Trigonometric Approximation}
{Trigonometric Approximation}
(vi) tan x ≈ x {Trigonometric Approximation}
Conceptual Note(s)
In Binomial Series, if x 1, then only the first two
terms are significant. It is so because the values of second and the higher order terms become very small
and hence can be neglected. So the following expressions can be re-written as
Further more, we have 1 = 1 and 0 = 1.
SERIES EXPANSIONS
{Exponential Series}
1
1
(d) ln ( 1 ± x ) = ± x − x 2 ± x 3 − 2
3
{Logarithmic Series}
So for A = 60° , we have
(
n ( n − 1) 2
x +
2!
(c) sin ( 300° )
Solution
⇒
1.9
( 1+ x )n = 1+ nx
⇒
( 1+ x )−n = 1− nx ⇒
( 1− x )n = 1− nx ⇒
( 1− x )−n = 1+ nx 11/28/2019 6:44:47 PM
1.10 JEE Advanced Physics: Mechanics – I
FUNCTION: AN INTRODUCTION
Illustration 9
13
Evaluate ( 1001 )
upto six places of decimal.
⇒
( 1001 )1 3 = 10 ⎡⎢ 1 + 0.00033 − ( 0.000001 ) + ..... ⎤⎥
A key idea in mathematical analysis and in Physics is
the idea of dependence of one quantity on the other.
A quantity depends on another if the variation of one
of them is accompanied by a variation of other. We
must have seen Mathematicians speaking about an
independent variable and the dependent variable.
Similarly in Physics, it is better to think in terms of
cause and effect or interdependent quantities.
We are aware of the fact that the area of a circle
depends upon its radius. Mathematically speaking,
the area of a circle is a function of its radius.
Similarly in Physics we observe that the volume
of a given mass of a gas at a fixed temperature is a
function of the pressure of the gas. That is we can say
that the cause of the change in temperature will produce an effect that produces the change in pressure
of the gas.
⇒
( 1001 )1 3 = 10 ( 1.0003301 ) = 10.003301 (Approx.)
REPRESENTATION OF A FUNCTION
Solution
( 1001 )1 3 = ( 1000 + 1 )1 3 = 10 ( 1 + 0.001 )1 3
By comparing the given equation with standard
equation
(1 + x )n = 1 + nx +
⇒
n(n − 1) 2
x + ......
2!
x = 0.001 and n =
1
3
⎡ 1
⎛ Neglected ⎞ ⎤
= 10 ⎢ 1 + ( 0.001 ) + ⎜
⎥
⎝ Terms ⎟⎠ ⎦
⎣ 3
⇒ 10 ( 1 + 0.001 )
13
1
9
⎣
⎦
Illustration 10
The value of acceleration due to gravity ( g h ) at
a height h above the surface of earth is given by
gR2
gh =
. Find the approximate value of g h
( R + h )2
when h R.
Solution
⎛ R ⎞
gh = g ⎜
⎝ R + h ⎟⎠
2
2
⎛ 1 ⎞
⎜
h⎟
⎜⎝ 1 + ⎟⎠
R ⇒
gh = g
⇒
h⎞
⎛
gh = ⎜ 1 + ⎟
⎝
R⎠
⇒
2
⎡
⎤
h ( −2 ) ( −3 ) ⎛ h ⎞
g h = g ⎢ 1 + ( −2 ) +
⎜⎝ ⎟⎠ + ..... ⎥
2!
R
R
⎣
⎦
⇒
2h ⎞
⎛
gh = g ⎜ 1 − ⎟
⎝
R⎠
01_Mathematical Physics_Part 1.indd 10
−2
A function is denoted by symbols like of f ( x ) ,
F ( x ) , ϕ ( x ) … . and is read as function of x . Thus if
y is a function of x , we may write y = f ( x ) .
Conceptual Note(s)
It must be clearly understood that f(x) does not mean
f into x, but is only a symbolic way of representing
some function of x.
The dependence of one quantity on another can be
quantitatively expressed in three different ways:
(a) Tabular Presentation
(b) Graphical Presentation
(c) Mathematical Presentation/Equations
Tabular Presentation
Let us consider the distance covered by an automobile, moving at constant speed, as a function of time.
Data for a particular example of such motion may be
presented numerically, as in the following Table. The
exact mathematical relationship between the time
and the distance in this example is not immediately
obvious while examining the table. This is one of the
11/28/2019 6:44:51 PM
Chapter 1: Mathematical Physics
1.11
disadvantages of tabular presentation. Although the
numerical values can be precisely specified, they do
not at once convey the clear picture of the dependence
of variables on one another. This can be better visualised by drawing the graph for the varying quantities.
This equation can also be expressed as
Table 1.2
The equation provides the most concise expression of
a functional relationship.
Time and Distance for a moving Automobile
Elapsed Time (min)
Distance (km)
0
0
2
1.5
4
3.0
6
4.5
8
6.0
10
7.5
Graphical Presentation
Let us plot the same data on a graph as shown in the
figure.
Distance, s(km)
9
6
where v0 is a constant whose value in this example is
v0 = 0.75 kmmin -1
SLOPE OF A LINE
The slope of a line in a graph is defined as the tangent of the angle (measured in anticlockwise direction) that the line makes with the positive direction
of the horizontal axis. This angle is designated by θ .
From the distance vs time plot shown in “Graphical
Presentation”, we have
s
= v0
t
is the slope of the straight line that has been plotted
for the variation of the distance with time.
Here we observe that the quantity tan θ is a
dimensional quantity. We always measure slope as a
vertical increment divided by a horizontal increment
on a graph, each increment being measured in the
appropriate unit for the quantity in the problem.
tan θ =
Illustration 11
If y = f ( x ) = x 2 − 3 x + 5 , find f ( 0 ) and f ( 1 ) .
3
θ
2 4 6 8 10 Time, t(min)
Variation of distance with time
Here we plot the Time (an independent variable) horizontally and the Distance (a dependent variable) vertically. Each pair of numbers in the Table 10.2 gives a
single point on the graph. It is immediately seen that
the points when joined give us a straight line.
Mathematical Presentation/Equations
The equation that fits the above tabular and graphical
data is
s = 0.75t
Where s represents the distance in kilometre and t
represents the time in minutes.
01_Mathematical Physics_Part 1.indd 11
s = v0 t
Solution
To find the value of a function at a particular value of
the independent variable, let us put the given value
in the expression of the function and simplify. The
result followings.
Here, f ( x ) = x 2 − 3 x + 5
f (0) = (0) − 3(0)+ 5 = 5
2
f (1) = (1) − 3 (1) + 5 = 3
2
CONCEPT OF LIMIT OF FUNCTIONS:
MEANING OF THE SYMBOL x → a
When the independent variable is gradually taken to
a definite value, say a, the dependent variable, i.e.,
the function will lead to another definite value, say l.
This value is defined as the limiting value of the
11/28/2019 6:44:54 PM
1.12 JEE Advanced Physics: Mechanics – I
function as the independent variable approaches the
given value a .
The arrow in the above symbol stands for gradual approach of x to a and the symbol is read as x
tending to a . If y = f ( x ) approaches a value l , as x
approaches a , we say that the limiting value of f ( x )
is l when x approaches a and this is symbolically
written as
When x approaches 2 from left, we observe f(x)
approaches 3 from the left.
lim f (x ) = l
x→ a
This symbol is read as the limit of the function is l
when x tends to a.
n=8
n→∞
As we increase the number of sides, the sides will be
shorter and shorter in size, the area of the polygon
will increase and ultimately when n is made infinitely
large, the area of the polygon will become equal to
the area of circle. Thus we may say that the limiting
value of the area of a polygon of n sides inscribed
in a circle is the area of the circle itself, as n tends to
infinity. ( Just think that the circle is a polygon having
infinite number of sides)
Mathematically, lim Area = π a2
n→∞
(b) Let us consider the function y = f ( x ) = x 2 − 1. Let us
find lim f ( x ).
x →2
01_Mathematical Physics_Part 1.indd 12
1.9
2.6100
1.99
2.96010
1.999
2.996001
1.9999
2.99960
x
y = f(x) = x2 - 1
2.2
3.84000
2.1
3.41000
2.01
3.04010
2.001
3.004
2.0001
3.00040
From the above two tables, it is clear that as x tends
to 2, y = f ( x ) = x 2 − 1 approaches or tends to 3
n = 10
y = f(x) = x2 - 1
When x approaches 2 from right, we observe f(x)
approaches 3 from the right.
EXAMPLES
(a) Let us inscribe a polygon of n sides in a circle of
radius a. The area A of the polygon will depend on
the number of sides n. Hence A = f ( n ) .
n=6
x
⇒
lim x 2 − 1 = 3
x →2
The process of finding the limiting value of a function
in the above way is actually the fundamental way of
finding the limiting value of a function called the first
principle. But this is considered to be a lengthy process. Every time it is not possible to find the limiting value by this process and that too for all kinds of
functions.
Simply by putting x = 2 in the above expression
will give us the limiting value. Hence the shortcut
method of finding the limiting value of a function is
to make direct substitution of the limiting the value
of the independent variable in the expression of the
function. But this technique may not work for all
kinds of functions. Actually we have learnt the concept of limit in an easy manner, but this concept is
useful whenever we have to find the value of the
functions at a point where they do not have indeterminate values. Consider another example to have a
clear concept.
11/28/2019 6:44:58 PM
Chapter 1: Mathematical Physics
(c) Let us take the function
y = f(x)=
From the above two tables, it is clear that as x tends
x2 − 9
to 3, y = f ( x ) =
approaches (or tends to) 6
x −3
x2 − 9
x −3
Here the student generally concludes that the above
expression could also have been easily written as
y = f ( x ) = x + 3 . But this is only true when x ≠ 3.
Actually we shall not be able to find the value of this
function at x = 3, because at x = 3 the function has
0
a value
which is an indeterminate value. For this
0
let us again consider the values of x approaching
from the left as well as the right towards 3.
y=
x2 – 9
x–3
(0, 6)
(0, 3)
(–3, 0) O
1.13
(3, 0)
x
When x approaches 3 from left, we observe f(x)
approaches 6 from the left.
x2 − 9
x−3
⇒
x2 − 9
=6
x →3 x − 3
lim
Conceptual Note(s)
x2 − 9
x −3
is discontinuous at x = 3. A function is said to
be continuous when you can draw it in one go
with your pen without its tip losing contact with
the paper. However, if we consider the function
y = x + 3, we observe that it is continuous.
(b) Remember that you must not cancel the common
factor from the numerator and the denominator
till you are sure that the common factor is non
zero.
x 3 y4
So, the expression
= x 2 y 2 only if x ≠ 0 and
xy 2
y ≠ 0.
(c) A very important formula of limit extensively used
in Physics is
sinϕ
lim
=1
ϕ →0 ϕ
(a) Please note here that the function y =
x
y = f(x) =
2.9
5.9
2.99
5.99
DERIVATIVE OF A FUNCTION
2.999
2.9999
Meaning of Dx
5.999
5.9999
For a finite but small increment in x , we use symbol
Δx . Please note that this is also not to be read as Δ
multiplied by x . It stands for a small but finite increment in x and is treated as a single quantity. This
should be read as ‘delta x ’.
When x approaches 3 from right, we observe f(x)
approaches 6 from the right.
x2 − 9
x−3
x
y = f(x) =
3.1
6.1
3.01
6.01
3.001
3.0001
6.001
6.0001
01_Mathematical Physics_Part 1.indd 13
Meaning of dx
Whenever a variable is changed by an infinitesimal
(extremely small) amount, then that change is called
the differential of the variable. This is denoted by dx
and read as ‘dee x ’. Again please do not misinterpret this as d into x. dx is merely a r­epresentation
standing for a very, very small increment in x. This
11/28/2019 6:45:02 PM
1.14 JEE Advanced Physics: Mechanics – I
should be treated as a single quantity just as sin θ
which again is not the product of sine and θ .
Slope has a simple physical meaning. It is the
rate of change of the quantity being plotted vertically with respect to the quantity being plotted horizontally or we can say that it is the rate of change of
dependent variable with respect to an independent
variable. Mathematically, derivative of the function
gives the instantaneous slope of that function.
(b) Instantaneous slope is at a point or an instant and
it is the slope of the tangent drawn to the curve at
that point.
y
Tangent
P
θ
dx
Slope and Derivative of a Function
Slope has a simple physical meaning. It is the rate of
change of the quantity being plotted vertically with
respect to the quantity being plotted horizontally or
we can say that it is the rate of change of dependent variable with respect to an independent variable.
Mathematically,
dy
(derivative of the function y w.r.t. x) gives
dx
the instantaneous slope of the function at a
dy
point.
is also called as the Rate Measurer.
dx
Δy
(b)
gives the average slope of a curve y between
Δx
two points.
dy
O
x
So, instantaneous slope of the curve at the point P
is
⎛ of angle which tangent to ⎞
dy
= tanθ = tan ⎜ point P makes with ⎟
⎜
⎟
dx
+ x direction
⎝
⎠
(a)
Conceptual Note(s)
Please note that:
(a) Average slope is always in an interval i.e. between
two points and it is the slope of the chord that
joins the two points.
y
A
O
Chord
B
θ
Δy
Δx
+x
x
So, average slope of the curve between the points
A and B is
⎛ of angle which chord ⎞
Δy
= tanθ = tan ⎜ joining points A and B ⎟
⎜
⎟
Δx
⎝ makees with + x direction ⎠
01_Mathematical Physics_Part 1.indd 14
DEFINITION OF DIFFERENTIAL
COEFFICIENT
Consider a function y = f ( x ) . Let the value of x
changes to x + Δx . Correspondingly the value of y
changes from y to y + Δy . The limiting value of the
Δy
ratio
when Δx tends to zero is called the differΔx
ential coefficient of y with respect to x . It is denoted
dy
by the symbol
.
dx
dy
Δy
= lim
dx Δx →0 Δx
The process of finding the differential coefficient of a
function is called differentiation or the derivative of
the function and this signifies the instantaneous slope
of that function.
Thus,
MATHEMATICAL DEFINITION
Consider a function y = f ( x ) . Let the value of x
changes to x + Δx . Correspondingly the value of y
changes from y to y + Δy . The limiting value of the
Δy
ratio
when Δx tends to zero is called the differΔx
ential coefficient of y with respect to x . It is denoted
dy
by the symbol
.
dx
11/28/2019 6:45:06 PM
Chapter 1: Mathematical Physics
Thus,
So,
dy
Δy
= lim
dx Δx →0 Δx
\
dy
d
=
⎡ f ( x ) ⎤⎦ = f ’ ( x )
dx dx ⎣
RULES OF DIFFERENTIATION
then, y + Δy = f ( x + Δx )
⇒
y + Δy − y = f ( x + Δx ) − f ( x )
⇒
Δy = f ( x + Δx ) − f ( x )
⇒
Δy f ( x + Δx ) − f ( x )
=
Δx
Δx
The process of finding the derivative of a function is
called differentiating the function.
Differentiation obeys several simple rules that are
worth Committing To Memory (CTM).
RULE 1:
f ( x + Δx ) − f ( x )
dy
= f ′ ( x ) = lim
Δx → 0
dx
Δx
The derivative of f ( x ) at x = a is denoted by f ′ ( a ) .
In other words the slope of the function y = f ( x ) at
x = a is given by f ′ ( a ) .
GEOMETRICAL INTERPRETATION OF
DERIVATIVE
Let us consider the graph of y = f ( x ) as shown in
figure.
Tangent at point P
y
Q
T′ Δy = f(x + Δx) – f(x)
P
R
ϕ
T
O
L
x
x + Δx
Δx
M
∠QPR = θ
∠PTL = ϕ
x
Let P and Q be the two points on it. Then
PR = LM = Δx
QR = Δy
Δ y f (x + Δx)− f (x)
tan θ =
=
Δx
Δx
The average slope of the curve in the internal PQ .
As Q → P along the curve, Δx → 0 , θ → ϕ and PQ
becomes tangent TPT ′ at P .
01_Mathematical Physics_Part 1.indd 15
θ →ϕ
d ( n ) dy
x =
at P dx
dx
dy
⎛
⎞
tan ϕ ⎜ i.e.,
at P ⎟ is the slope of tangent at P.
⎝
⎠
dx
If y = f ( x )
⇒
tan ϕ = lim tan θ =
1.15
Derivative of a constant is zero.
d
( constant ) = 0
dx
RULE 2:
The derivative of a constant times a function is the
constant times the derivative of the function.
d
du
( au ) = a
dx
dx
RULE 3:
The derivative of the sum of the functions is the sum
of their derivatives.
d
du dv dw
( u ± v ± w ± .... ) =
±
±
± .....
dx
dx dx dx
RULE 4: PRODUCT RULE
Derivative of product of two functions is given as
d
dv
du
( uv ) = u + v
dx
dx
dx
d
d
d
1 × 2 ) =1 2 + 2 1
(
dx
dx
dx
RULE 5: QUOTIENT RULE
Derivative of a quotient is given as
d ⎛ u⎞
⎜ ⎟=
dx ⎝ v ⎠
v
du
dv
−u
dx
dx
v2
11/28/2019 6:45:12 PM
1.16 JEE Advanced Physics: Mechanics – I
d
d
(Den r ) (Num r ) − (Num r ) (Den r )
d ⎛ Num r ⎞
dx
dx
⎜
⎟=
dx ⎝ Den r ⎠
( Den r )2
RULE 6: CHAIN RULE
Suppose f is a function of u , which in turn is a funcdf
tion of x . The derivative
can be written as the
dx
product of two derivatives.
df df du
=
×
dx du dx
IMPORTANT DIFFERENTIAL FORMULAE
d
( constant ) = Zero
dx
d
[ cos ( ax + b ) ] = − a sin ( ax + b )
dx
d
du dv dw
(u ± v ± w) =
±
±
dx
dx dx dx
d
[ tan ( ax + b ) ] = a sec2 ( ax + b )
dx
d ( n)
x = n x n−1
dx
d
[ sec ( ax + b ) ] = a sec ( ax + b ) tan ( ax + b )
dx
d ( n)
kx = k ( nx n−1 )
dx
d
[ cot ( ax + b ) ] = − a cosec2 ( ax + b )
dx
d
dv
du
( uv ) = u + v
dx
dx
dx
d
[ cosec ( ax + b ) ] = − a cosec ( ax + b ) cot ( ax + b )
dx
d
dw
dv
du
( uvw ) = uv
+ uw + vw
dx
dx
dx
dx
d
1
( log e x ) = x
dx
dv
du
u
d ⎛ u ⎞ v dx − dx
⎜ ⎟=
dx ⎝ v ⎠
v2
d
a
⎡ log e ( ax + b ) ⎤⎦ =
dx ⎣
ax + b
( Den r ) d ( Numr ) − ( Numr ) d ( Denr )
d ⎛ Num r ⎞
dx
dx
⎜
⎟=
dx ⎝ Den r ⎠
( Den r )2
d ( x) x
e =e
dx
If y is a function of u and u is a function of x, then
dy dy du
=
×
dx du dx
d ( x) x
a = a log e a
dx
If y is a function of u, u a function of u, v a function of w
dy dy du dv dw
and w a function of x, then
=
×
×
×
dx du dv dw dx
d ( kx )
e = ke kx
dx
d
( sin x ) = cos x
dx
d ( kx )
a = kakx log e a
dx
d
( cos x ) = − sin x
dx
d
⎡ log e ( sec x + tan x ) ⎤⎦ = sec x
dx ⎣
01_Mathematical Physics_Part 1.indd 16
11/28/2019 6:45:17 PM
Chapter 1: Mathematical Physics
d
( tan x ) = sec2 x
dx
d
⎡ log e ( cot x + cosec x ) ⎤⎦ = − cosec x
dx ⎣
d
( cot x ) = − cosec2 x
dx
d
( x log e x − x ) = log e x
dx
d
( sec x ) = sec x tan x
dx
d
⎡ log e ( cos x ) ⎤⎦ = − tan x
dx ⎣
d
( cosec x ) = − cosec x cot x
dx
d
⎡ log e ( sin x ) ⎤⎦ = cot x
dx ⎣
d
[ sin ( ax + b ) ] = a cos ( ax + b )
dx
n
n −1
d ( 2
ax + bx + c ) = ( 2 ax + b ) ⎡⎣ n ( ax 2 + bx + c ) ⎤⎦
dx
Illustration 12
dy
⇒ = ( cos u )( 2x ) = 2x cos u
dx
dy
⇒
= 2x cos ( x 2 )
dx
2
Find the derivative of y = 3 x .
Solution
dy
d
= 3 ( x 2 ) = 3 ( 2x ) = 6 x
dx
dx
{
d ( n)
∵
x = nx n −1
dx
}
Since the derivative of the sum is the sum of the
derivatives, so
(using rule 3)
Illustration 14
Find the derivative of y = sin ( x 2 ) .
Solution
Let us assume u = x 2 , then y = sin u.
dy
du
= cos u and
= 2x
du
dx
dy dy du
Since,
=
(∵ of chain rule)
dx du dx
Then
01_Mathematical Physics_Part 1.indd 17
dy
.
dx
dy
d
( sin x + cos x )
=
dx dx
Using RULE 3, we get
⇒
Solution
⇒
If y = sin x + cos x , then find
Since y = sin x + cos x
Find the derivative of y = x 3 + 3 x 2 .
⇒
Illustration 15
Solution
Illustration 13
dy
d ( 3
x + 3x 2 )
=
dx dx
dy
d ( 3) d ( 2)
3x =
x +
dx dx
dx
dy
= 3x 2 + 6x
dx
1.17
⇒
dy
d
d
=
( sin x ) + ( cos x )
dx dx
dx
dy
= cos x − sin x
dx
Illustration 16
Find the derivative of y = x sin x.
Solution
dy
d
⎛ dx ⎞
= x cos x + sin x = x ( sin x ) + ( sin x ) ⎜
⎝ dx ⎟⎠
dx
dx
(using product rule)
Illustration 17
Differentiate the following w.r.t. x.
(a) sin x − cos x
(b) sin x + e x
11/28/2019 6:45:24 PM
1.18 JEE Advanced Physics: Mechanics – I
Solution
(a)
(b)
(d)
d
d
d
( sin x − cos x ) = ( sin x ) − ( cos x )
dx
dx
dx
d
( sin x − cos x ) = cos x + sin x
⇒
dx
(e) Let y = 6 log e x − x − 7
dy
d
=
6 log e x − x − 7
dx dx
dy
d
d
d
= 6 ( log e x ) − ( x1/2 ) − ( 7 )
⇒
dx
dx
dx
dx
dy 6
1
= −
⇒
dx x 2 x
Illustration 18
x2 + 1
.
x
⇒
Differentiate the following w.r.t. t.
x
(
)
(
)
(
d 2
⎡ d
⎤
x + 1 − ⎢ (x) ⎥ x 2 + 1
dx
⎣ dx
⎦
x2
(a) sin ( t 2 ) )
Illustration 19
dy
If y = x , then find
.
dx
Solution
Given y =
Solution
(a)
d(
d
sin t 2 ) = cos t 2 ( t 2 ) = 2t cos t 2
dt
dt
(b)
d ( sin t ) sin t d
( sin t ) = e sin t .cos t
e
=e
dt
dt
(c)
d
[ sin ( ω t + θ ) ] = cos ( ω t + θ ) d ( ω t + θ )
dt
dt
1
x = x2
⇒
{
1
1
dy d ⎛ 2 ⎞ 1 2 −1
1
d
∵ ( x n ) = nx n −1
=
= ⎜x ⎟= x
dx
dx dx ⎝ ⎠ 2
2 x
}
d
[ sin ( ω t + θ ) ] = ω cos ( ω t + θ )
dt
Illustration 22
Differentiate
Illustration 20
Differentiate the following w.r.t. x.
3
(d) 2x 3 − e x
(b)
x
(c) ax + bx + c
(e) 6 log e x − x − 7
d ( 3)
x = 3x 2
dx
1
d
1
1
1
( x )1/2 = ( x ) 2 −1 = ( x )−1/2 =
(b)
dx
2
2
2 x
(a)
d ( 2
d
d
d
ax + bx + c ) = a ( x 2 ) + b ( x ) + ( c )
dx
dx
dx
dx
d ( 2
⇒
ax + bx + c ) = 2 ax + b
dx
01_Mathematical Physics_Part 1.indd 18
x2 + ex
w.r.t. x.
log x + 20
Solution
2
Solution
(c)
(b) e sin t
(c) sin(ω t + θ )
2
2
dy 2x − x + 1 x 2 − 1
=
= 2 (using quotient rule)
x2
x
dx
(a) x
)
Illustration 21
Solution
dy
=
dx
(
⇒
d (
d
d
( sin x ) + ( e x ) = cos x + e x
sin x + e x ) =
dx
dx
dx
Find the derivative of y =
d ( 3
d
d
2x − e x ) = 2 ( x 3 ) − ( e x ) = 6 x 2 − e x
dx
dx
dx
Let y =
Then
x2 + ex
log x + 20
dy
d ⎛ x2 + ex ⎞
=
dx dx ⎜⎝ log x + 20 ⎟⎠
(
) (
)
d 2 x
2
x d
dy ( log x + 20 ) dx x + e − x + e dx ( log x + 20 )
=
dx
( log x + 20 )2
⇒
dy
=
dx
1
( log x + 20 ) ( 2x + e x ) − ( x 2 + e x ) ⎛⎜⎝ x + 0 ⎞⎟⎠
( log x + 20 )2
11/28/2019 6:45:30 PM
1.19
Chapter 1: Mathematical Physics
Illustration 23
Illustration 26
dy
If y = a sin(ω t), where a and w are constants, find
.
dt
If y = x 2 sin x , find
Solution
Solution
Given y = a sin ω t
⇒
⇒
⇒
dy d
d
= ( a sin ω t ) = a ( sin ω t )
dt dt
dt
dy
d
= a cos ( ω t ) ( ω t ) (∵ of chain rule)
dt
dt
dy
= aω cos ( ω t )
dt
Illustration 24
1
x
2
− 3 log e x at x =
π
.
2
Solution
1
− 3 log e x
x2
Slope of the function is the derivative of the function.
So,
dy
d
d
d 1
=
( sin x ) + ⎛⎜⎝ 2 ⎞⎟⎠ − 3 ( log e x )
dx
dx dx
dx x
We have y = sin x +
dy
2 3
= cos x − 3 −
dx
x
x
π
Slope at x =
is given by
2
2
3
⎛ dy ⎞
⎛π⎞
−
⎜⎝
⎟⎠ π = cos ⎜⎝ 2 ⎟⎠ −
3
dx x =
(π 3 ) (π 2 )
⇒
2
⇒
Given y = x 2 sin x
16 6
16 6
⎛ dy ⎞
⎜⎝
⎟⎠ π = 0 − 3 − = − 3 −
dx x =
π
π
π
π
2
Illustration 25
dy
.
dx
Solution
⇒
dy
d
d 2
= x 2 × ( sin x ) + sin x ×
x
dx
dx
dx
(using product rule)
dy
= x 2 cos x + 2x sin x.
dx
Illustration 27
Find the slope of the tangent to the curve y = 3 x 2 − 5
at the point ( 2, 7 ) .
Solution
We have
y = 3x 2 − 5
⇒
dy
= 6x
dx
At point ( 2, 7 ) , we have
⇒
dy
= tan θ = 6 ( 2 ) = 12
dx
tan θ = 12
Illustration 28
Find the inclination with the x-axis of the tangent to
the curve y 2 = 4 x at ( 1, 2 ) .
Given y 2 = 4 x
( )
( )
dy
d
d
d
=
3e x −
5x 3 + ( 3 ) (∵ of rule 3)
dx dx
dx
dx
dy
= 3 e x − 5 3 x 2 + 0 = 3 e x − 15x 2 . dx
dy
= 4
dx
⇒
2y
⇒
dy 2
=
dx y
Given y = 3 e − 5x + 3
01_Mathematical Physics_Part 1.indd 19
( )
3
x
⇒
)
dy
d 2
x sin x
=
dx dx
Solution
If y = 3 e x − 5x 3 + 3 , then find
⇒
(
⇒
⇒
Find the slope of the function
y = sin x +
dy
.
dx
(taking derivative of both sides)
( )
11/28/2019 6:45:35 PM
1.20 JEE Advanced Physics: Mechanics – I
At the point ( 1, 2 )
dy 2
= =1
dx 2
⇒ tan θ = 1 ⇒ q = 45°
2
⇒
dy
1 − 2( 2 )
1− 8 7
=
=
=
dx x = 2 , y = −5 2 − 2 ( 5 ) 2 − 10 8
and
dy
1 − 2 ( −3 )
1 − 18
17
=
=
=−
1+ 6
7
dx x = 2 , y = −3 1 − 2 ( −3 )
2
Illustration 29
APPLICATIONS OF DERIVATIVE
dy
2
Find
for y =
+ x x.
dx
x
With the help of differentiation, we can
Solution
⇒
⇒
(
1
3
(a) check whether the function is increasing or
decreasing.
(b) find the maximum and minimum value(s) of a
function.
(c) calculate the rate of change of quantity (say y)
w.r.t. another quantity (say x).
)
−
dy
d
=
2x 2 + x 2
dx dx
1
3
−
−
1
dy
3 −1
⎛ 1⎞
= 2⎜ − ⎟ x 2 + x2
⎝ 2⎠
2
dx
dy
1
3
=− 3 +
dx
2 x
x2
Increasing and Decreasing Function
Illustration 30
If we have 2 y − y 2 − x + x 3 + 9 = 0, then calculate
at x = 2.
dy
dx
Solution
Let us firstly calculate value of y, when x equals 2.
For this, we have
2y − y 2 − 2 + 8 + 9 = 0
⇒
2 y − y 2 + 15 = 0
⇒
f ( x1 ) < f ( x2 ) then f ( x ) is increasing if x1 < x2
⇒
f ( x1 ) > f ( x2 ) then f ( x ) is decreasing
As shown in the figure, when f ( x ) is increasing, the
tangent to the curve at any point, say P , makes an
acute angle with positive x-axis. The slope of the tangent is positive.
tan θ =
Thus,
2
⇒
y − 2 y − 15 = 0
⇒
y 2 − 5 y + 3 y − 15 = 0
⇒
( y − 5 )( y + 3 ) = 0
⇒
y = 5, y = −3
So, we are to calculate
A function f ( x ) is said to be increasing if f ( x )
increases as x increases, and decreasing if f ( x )
decreases as x increases.
In other words, if x1 < x2
dy
at ( 2, 5 ) and ( 2, − 3 ) .
dx
dy
>0
dx
As shown in the figure, when f ( x ) is decreasing,
the tangent to the curve at any point, say P , makes
an obtuse angle with positive x-axis. The slope of the
tangent is negative.
y
Now, 2 y − y 2 − x + x 3 + 9 = 0
⇒
⇒
⇒
dy
dy
− 2y
− 1 + 3x 2 = 0
dx
dx
dy
dy
2
− 2y
= 1 − 3x 2
dx
dx
Q
P
2
dy 1 − 2x 2
=
dx 2 − 2 y
01_Mathematical Physics_Part 1.indd 20
θobtuse
Thus, tan θ =
θ acute
x
dy
<0
dx
11/28/2019 6:45:42 PM
Chapter 1: Mathematical Physics
Maximum and Minimum Values
of a Function
Let us suppose a quantity y depends upon another
quantity x in a manner shown in the figure. It becomes
maximum at x1 and minimum at x2 . At those points,
the tangent to the curve is parallel to the x-axis and
hence its slope is tan θ = 0 .
dy
dx
So, for the maximum or minimum value of y , we
dy
have
=0
dx
Just before the maximum point, the slope is positive.
At the maximum point, it is zero And just after the
dy
maximum point, it is negative. Thus,
decreases at
dx
dy
the maximum point, i.e., the rate of change of
is
dx
negative at the maximum point,
d2 y
>0
dx 2
As shown in the figure, at the point of maximum and
minimum of a function the slope of the tangent at the
point is zero.
Thus, tan θ =
But the slope of curve =
So
⇒
Conceptual Note(s)
(a) f ( x ) is maximum at a point x = a, if
(i) f ′ ( a ) = 0 and
(ii) f ′ ( x ) changes in sign from positive to negative when x passes through the point x = a.
In other words, the second derivative of the
function at x = a is negative
i.e., f ′′ ( a ) < 0.
(i) f ′ ( a ) = 0 and
(ii) f ′ ( x ) changes in sign from negative to positive when x passes through the point x = a.
In other words, the second derivative of the
function at x = a is positive.
d2 y
< 0
2
y
P
i.e., f ′′ ( a ) > 0.
x
O
Q
The maximum and minimum of a function
Hence, condition for the maximum value of y is
dy
= 0 and
dx
d2 y
< 0
dx 2
Similarly, at a minimum point, the slope changes
from negative to positive. The slope increases at such
d ⎛ dy ⎞
>0
a point. Hence,
dx ⎜⎝ dx ⎟⎠
Hence, condition for the minimum value of y is
dy
= 0 and
dx
01_Mathematical Physics_Part 1.indd 21
dy
=0
dx
(b) f ( x ) is minimum at a point x = a, if
d ⎛ dy ⎞
<0
dx ⎜⎝ dx ⎟⎠
dx
1.21
Illustration 31
Consider a function y = sin x + cos x . Find the maximum value of the function.
Solution
y = sin x + cos x
dy
= cos x − sin x
dx
For a function to be a MAXIMUM
⇒
⇒
dy
=0
dx
cos x − sin x = 0
⇒
tan x = 1
⇒
⇒ α=
π
4
11/28/2019 6:45:48 PM
1.22 JEE Advanced Physics: Mechanics – I
Verification
2
d y
dx
⇒
2
Once again differentiating w.r.t. time t, we get
= − sin x − cos x
d2 y
dx
2
π
at x =
4
Since, at x =
= − sin
d2 x
π
π
− cos = − 2
4
4
π d2 y
,
<0
4 dx 2
So y is MAXIMUM at x =
π
4
y = sin x + cos x
⎛π⎞
⎛π⎞
y at x = π = sin ⎜ ⎟ + cos ⎜ ⎟ = 2
⎝
⎠
⎝ 4⎠
4
4
Illustration 32
The height reached in time t by a particle thrown
1
upward with a speed u is given by h = ut − gt 2 .
2
Find the time taken in reaching the maximum height.
Solution
dh
=0
dt
2 gt
d ⎡
1
⎤
ut − gt 2 ⎥ = u −
=0
⎢
dt ⎣
2
⎦
2
u
t=
g
For maximum height
⇒
= 6t − 6
At t = 0 ,
d2 x
dt 2
At t = 2 s ,
Maximum Value
⇒
dt 2
Illustration 33
The distance travelled by a body as a function of time
is given by x = t 3 − 3t 2 + 6, where x is m and t is
in S . Find the maximum and minimum displacement of the body from the origin. Also find the time
at which it occurs.
Solution
= −6 < 0 (maximum)
d2 x
= 6 > 0 (minimum)
dt 2
The maximum displacement occurs at t = 0 and is
equal to
( x )max = ( 0 )3 − 3 ( 0 )2 + 6 = 6 m
The minimum displacement occurs at t = 2 s and is
equal to
( x )min = ( 2 )3 − 3 ( 2 )2 + 6 = 2 m
dy
as Rate Measure
dx
The rate of change of a quantity ( y ) with respect
to another quantity ( x ) is defined as the ratio of
the change of y to change is x , however small the
change in x may be
If Δy be the change in y corresponding to a
change Δx is x , then according to the definition, the
rate of change of y with respect to x is the limiting
Δy
value of the ratio
when Δx tends to zero.
Δx
So, rate of change of y with respect to x is
dy
Δy
= lim
Δ
x
→
0
dx
Δx
Conceptual Note(s)
When we simply say rate of change y, we mean change
dy
of y with respect to time. So, the rate of change of y is .
dt
Differentiating x with respect to t , we get
dx
= 3t 2 − 6t
dt
dx
=0
For maximum and minimum displacements
dt
∴ 3t 2 − 6t = 0 ⇒
⇒
3t ( t − 2 ) = 0 t = 0 and t = 2 s 01_Mathematical Physics_Part 1.indd 22
Illustration 34
The area of a blot of ink is growing such that after
t second, its area is given by A = ( 3t 2 + 7 ) cm 2 .
Calculate the rate of increase of area at t = 5 seconds.
Solution
Given A = 3t 2 + 7
11/28/2019 6:45:55 PM
Chapter 1: Mathematical Physics
\
⇒
Differentiating both sides w.r.t. time
dA
= 6t
dt
⎛ dA ⎞
= 6 × 5 = 30 cms −2
⎜⎝
⎟
dt ⎠ t = 5
dV d ⎛ 4
⎞
= ⎜ π R3 ⎟
⎝
⎠
dt dt 3
Illustration 35
A metal ring is being heated so that at any
instant of time t in second, its area is given by
t
A = 3t 2 + + 2 m 2 . What will be the rate of increase
3
of area at t = 10 sec.
⇒
dV ⎛ 4 ⎞ ⎛ 2 dR ⎞
= ⎜ π ⎟ ⎜ 3R
⎟
dt ⎝ 3 ⎠ ⎝
dt ⎠
⇒
dV
dR
= 4π R2
dt
dt
at R = 1 cm and when
Rate of increase of area
dA d ⎛ 2 t
1
⎞
= ⎜ 3t + + 2 ⎟ = 6t +
⎝
⎠
dt dt
3
3
INTEGRATION: AN INTRODUCTION
1 181 2 −1
⎛ dA ⎞
= 6 × 10 + =
m s
⎜⎝
⎟⎠
dt t =10 sec
3
3
Illustration 36
Find the rate of change in area of a square of side 4 cm
when its side is increasing at the rate of 0.01 cm per
second.
Let A be the area of the square; a be the length of
the side
We have
A = a2
⇒
The word “integral” simply means “the whole”, and
the process of adding or summing up a large number of infinitesimal elements of a quantity is called
integration.
If x be supposed to be made up of a large number
of infinitesimal elements each equal to dx , it is obvious that if we add up all these dx together, we shall
get the total x . Mathematically we put it as
Solution
⇒
dR 1
= cms −1 , we have
dt 2
dV
1
2
= 4π × ( 1 ) × = 2π cm 3 s −1
dt
2
Solution
⇒
( )
⇒
I(x) =
dI
dx
∫ f ( x ) dx
In the above expression, f ( x ) is called the integrand.
The symbol,
Illustration 37
The radius of an air bubble is increasing at the rate of
1
cms −1 . Determine the rate of increase in its volume
2
when the radius is 1 cm .
Solution
01_Mathematical Physics_Part 1.indd 23
∫ dx = x
(read it as integral of dx equals x ). Integration is the
inverse operation of differentiation.
Integration of f ( x ) simply means finding the function I ( x ) whose derivative is equal to f ( x ) .
Mathematically, f ( x ) =
( )
dA d 2
d 2
da
=
a =
a ×
dt dt
da
dt
dA
da
= 2a ×
dt
dt
dA
= 2 × 4 × .01 = 0.8 cms −2
dt
Volume of the spherical bubble V =
1.23
∫ is the symbol of integration and dx
indicates the variable of integration.
The function I ( x ) is also known sometimes as the
anti derivative of f ( x ) .
“The result of an indefinite integral, when differentiated, will give you the function that has been
integrated.”
4
π R3
3
11/28/2019 6:45:59 PM
1.24 JEE Advanced Physics: Mechanics – I
Table 1.3
Some Indefinite Integrals
x n +1
n+1
∫
∫
dx
∫ x = ∫ x dx = ln x
x n dx =
(provided n ≠ –1)
dx
∫ sin ( ax ) dx = − a cos ( ax )
1
∫ cos ( ax ) dx = a sin ( ax )
1
∫ (a + bx) = − b(a + bx)
2
dx
1
−1 ⎛ x ⎞
dx
1
⎛ a+x⎞
2
1
⎛ x−a⎞
2
1
2
1
∫ cot ( ax ) dx = a ln(sin ( ax ))
∫ a − x = 2a ln ⎝⎜ a − x ⎟⎠ (a − x > 0)
2
2
dx
2
∫ x − a = 2a ln ⎜⎝ x + a ⎟⎠ (x − a > 0)
2
2
xdx
1
∫
2
1
x
sin ( 2ax )
4a
x
sin ( 2ax )
4a
∫ sin ( ax ) dx = 2 −
2
∫ x + a = ln ( x + x + a )
∫ cos ( ax ) dx = 2 −
∫ x − a = ln ( x + x − a )
∫ sin ( ax ) = ∫ cosec ( ax ) dx = − a cot ( ax )
2
2
2
2
dx
2
2
dx
2
xdx
2
2
dx
ax ⎞ ⎤
1
2
2
1⎛
2
2
2
1
∫ tan ( ax ) dx = a (tan ( ax )) − x
−1 ⎛ x ⎞ ⎞
∫ a − x dx = − 2 ⎜⎝ x a − x + a sin ⎜⎝ a ⎟⎠ ⎟⎠
2
⎛
∫ cos ( ax ) = ∫ sec ( ax ) dx = a tan ( ax )
2
2
2
1⎡
1
2
2
2
∫ x ±a = x ±a
π⎞⎤
1
⎛ x⎞
⎛ x⎞
= sin −1 ⎜ ⎟ = − cos −1 ⎜ ⎟ ( a2 − x 2 > 0)
⎝
⎠
⎝ a⎠
2
2
a
a −x
dx
⎛ ax
∫ cosec ( ax ) dx = a ln(cosec ( ax ) − cot ( ax )) = a ⎢⎣ ln ⎜⎝ tan 2 ⎟⎠ ⎥⎦
2
dx
2
⎡
∫ sec ( ax ) dx = a ln(sec ( ax ) + tan ( ax )) = a ln ⎢⎣ tan ⎜⎝ 2 + 4 ⎟⎠ ⎥⎦
2
∫ a ± x = ± 2 ln(a ± x )
2
1
∫ tan ( ax ) dx = − a ln(cos ( ax )) = a (sec ( ax ))
1
∫ a + x = a tan ⎜⎝ a ⎟⎠
2
cx
1
1
2
1
x
cx
∫ a + bx = b ln(a + bx)
dx
e ax
( ax − 1)
a2
∫ a + be = a − ac ln(a + be )
−1
dx
xe ax dx =
2
1
∫
1
∫ x ± a dx = 2 ⎡⎢⎣ x x ± a ± a ln(x + x ± a ) ⎤⎥⎦
∫ cot ( ax ) dx = − a (cot ( ax )) − x
∫ x x ± a dx = 3 ( x ± a )
∫
1
∫ tan ( ax ) dx = x(tan ( ax )) − 2a ln(1 + a x )
1
∫ cot ( ax ) dx = x(cot ( ax )) + 2a ln(1 + a x )
x a2 − x 2 dx = −
2
2
2
)
2
2
1
1 ax
∫ e dx = a e
1
∫ e dx = − a e
ax
(
− ax
− ax
3
1 2
a − x2 2
3
2
2
2
3 2
2
2
2
2
∫
sin −1 ( ax ) dx = x(sin −1 ( ax )) +
cos −1 ( ax ) dx = x(cos −1 ( ax )) −
1 − a2x 2
a
1 − a2x 2
a
−1
−1
2 2
−1
−1
2 2
(an arbitrary constant should be added to each of these integrals)
01_Mathematical Physics_Part 1.indd 24
11/28/2019 6:46:04 PM
Chapter 1: Mathematical Physics
After each integral one must add a constant. The
­reason for adding a constant is given as follows:
The differential of x n+1 is ( n + 1 ) x n dx . The differential
of
(x
n+ 1
+c
) is ( n + 1 ) x dx because the differential
Illustration 41
Find
n
co-efficient of a constant is zero. Hence in general one
has to add a constant after performing an integration.
This constant is called the constant of integration.
∫ sin x dx.
2
Solution
Let I =
∫ sin x dx
2
Since, sin 2 x =
RULES FOR INTEGRATION
∫ c dx = c∫ dx , where c is a constant.
(b)
∫ ( u ± v ) dx = ∫ u dx ± ∫ v dx , where u and v are
(c)
∫ u dv = uv − ∫ v du
{ ∫
∵
∫
u dv + v du =
∫ d ( uv ) = uv }
∫ dx.
∫ ⎜⎝
x0 + 1
0+1
I = x + c , where c is a constant of integration.
∫
⇒
dx =
∫
( 1 ) dx =
∫
x 0 dx =
Illustration 39
∫ x dx
2
Evaluate
Solution
x 2+1 x 3
=
+ c , where c is a constant of
I = x dx =
2+1 3
integration.
∫
2
Illustration 40
1
∫ x dx
Evaluate
2
Solution
I=
dx
x2
=
∫
∫
⇒
I=
1⎛
dx − cos ( 2x ) dx ⎞
⎠
2⎝
∫
∫
sin ( 2x ) ⎞
1⎛
⎜⎝ x −
⎟⎠ + c
2
2
where c is a constant of integration.
⇒
I=
Illustration 42
Integrate the following w.r.t. x.
(a)
∫
(x )
3
1
+1
1
2
x2
=
x 2 dx =
1
+1
2
3
2
∫ cot x dx
⇒ I = ∫ ( cosec x − 1 ) dx
⇒ I = ∫ cosec x dx − ∫ dx
(b) I =
2
2
2
⇒
I = − cot x − x
1
∫ 1 − sin x dx
1
1 + sin x ⎞
⎛
⇒ I = ∫ ⎜⎝ 1 − sin x × 1 + sin x ⎟⎠ dx
1 + sin x
dx
⇒ I = ∫
1 − sin x
1
sin x
+
dx
⇒ I = ∫
cos x cos x
⇒ I = ∫ ( sec x + tan x sec x ) dx = tan x + sec x
(c) I =
2
2
1
x −2 + 1
x −2 dx =
= − +c
−2 + 1
x
01_Mathematical Physics_Part 1.indd 25
1
1 − sin x
Solution
Solution
Let I =
1
⎛ 1 − cos 2x ⎞
( 1 − cos 2x ) dx
⎟⎠ dx =
2
2
I=
(a) x1 2 (b) cot 2 x (c)
Illustration 38
Evaluate
1 − cos ( 2x )
2
⇒
(a)
the function of x .
1.25
2
2
11/28/2019 6:46:10 PM
1.26 JEE Advanced Physics: Mechanics – I
The constant of integration has disappeared during
the process of integrating a function within limits.
Illustration 43
Evaluate
∫ cos x dx
2
Solution
∫ cos x dx
1 + cos ( 2x ) ⎫
⎛ 1 + cos ( 2x ) ⎞
⎧
⇒ I= ⎜
∫ ⎝ 2 ⎟⎠ dx ⎨⎩∵ cos x = 2 ⎬⎭
2
Let I =
2
1 ⎡
dx + cos 2x ⎤⎥ dx
⎢
2 ⎣
⎦
∫ ∫
1
1⎛
sin ( 2x ) ⎞
⇒ I = ⎡⎢ dx + cos ( 2x ) dx ⎥⎤ = ⎜ x +
⎟⎠ + c
∫
2 ⎣∫
2
⎦ 2⎝
⇒
I=
Illustration 44
Evaluate
∫ e dx.
5x
Solution
Let I =
∫
e 5 x dx =
e 5x
+ c , where c is a constant of
5
GEOMETRICAL INTERPRETATION OF
DEFINITE INTEGRATION
As we have learnt, the graphical interpretation of differentiation is finding the slope of a curve. Integration
also has a simple graphical meaning. It is related to
finding the area under a curve.
If a function f(x) is expressed graphically in the form
f(x) vs x, the area under the curve between the limits
a and b means the area bounded by the curve of
f ( x ) , the x-axis and two lines x = a and x = b .
The area under the graph of a positive function is
defined to be positive. The area under (actually,
above) the graph of a negative function is defined to
be negative. As shown in the figure (c), positive and
negative area add algebraically and may even cancel.
The total area between definite limits of x is called
a definite integral. The notation for the definite integral is
b
Area = A =
integration.
∫ f ( x ) dx
a
DEFINITE INTEGRALS
f(x)
f(x)
When an integral is defined between two limits, it is
called a definite integral. The lower value of the limit
is called the lower limit and the higher value of the
limit is called the upper limit.
For example, if the integral of the function
f ( x ) is to be determined between the limits, x = a
and x = b, we represent it symbolically as
a
A>0
O
a
x
b
(a)
b
∫ f ( x ) dx
b
x
O
A<0
(b)
f(x)
a
where b is the upper limit and a is the lower limit.
We read it as “integral from a to b of the function
f ( x ) with respect to x ”.
If
∫ f ( x ) dx = ϕ ( x ) + c , then
b
∫ f ( x ) dx = ( ϕ ( x ) + c )
a
b
⇒
∫
b
A=0
a
b
O
x
(c)
a
Definition of the definite integral as area under
f ( x ) dx = ( ϕ ( b ) + c ) − ( ϕ ( a ) + c ) = ϕ ( b ) − ϕ ( a )
the curve. Area above the x-axis is defined to be
positive, area below the x-axis is negative.
a
01_Mathematical Physics_Part 1.indd 26
11/28/2019 6:46:14 PM
Chapter 1: Mathematical Physics
Conceptual Note(s)
A definite integral between fixed limits is a fixed quantity, not a function. It has a specific numerical value
and generally has a unit, which need not be a unit of
area. Just as the idea of slope is generalized from its
purely g­ eometric meaning and acquires a unit determined by the quotient of the vertically plotted quantity and the horizontally plotted quantity, the idea of
areas is also generalized and acquires a unit determined by the product of the vertically and horizontally plotted quantities.
Illustration 47
b
Find
Solution
b
Let I =
b
⇒
x 2 dx
2
Solution
3
3
⎡ 3 3 23 ⎤
⎛ x3 ⎞
Let I = x dx = ⎜
=
− ⎥
⎝ 3 ⎟⎠ 2 ⎢ 3
2 ⎦
⎣
2
⇒
a
⎛ b⎞
I = log e b − log e a = log e ⎜ ⎟ ⎝ a⎠
The momentum p of a particle changes with time t
dp
according to the relation
= (10 + 2t). If the momendt
tum is zero at t = 0 , what will be the momentum at
t = 10 s?
π
2
p
10
0
0
∫ dp = ∫ ( 10 + 2t ) dt
(
)
10
p = 10t + t 2
⇒
p = ⎡ 10 ( 10 ) + ( 10 ) ⎤ − ⎡ 10 ( 0 ) + ( 0 ) ⎤
⎣
⎦ ⎣
⎦
π
2
⇒
p = 200 kgms −1
0
Illustration 49
Solution
0
dp = ( 10 + 2t ) dt
⇒
0
∫ sin θ dθ = ( − cosθ )
Given
dp
= ( 10 + 2t )
dt
⇒
∫ sin θ dθ.
π
2
Solution
⇒
Illustration 46
Let I =
∫
b
dr ⎛
⎞
= ⎜ log e r ⎟ ⎠ a
r ⎝
2
27 8 54 − 24 30
I=
− =
=
=5
3 2
6
6
Evaluate
I=
Illustration 48
3
∫
dr
∫r
a
Illustration 45
∫
dr
∫ r.
a
⇒
Evaluate
1.27
0
2
2
t
⇒
⎤
⎡
⎛π⎞
I = − ⎢ cos ⎜ ⎟ − cos 0 ⎥ = − [ 0 − 1 ] = 1
⎠
⎝
2
⎦
⎣
01_Mathematical Physics_Part 1.indd 27
Evaluate the integral
are constants.
∫ A sin ( ωt ) dt, where A and ω
0
11/28/2019 6:46:18 PM
1.28 JEE Advanced Physics: Mechanics – I
Solution
t
Let I =
∫ A sin ( ωt ) dt
0
⇒
⎡ cos ( ω t ) t ⎤
I = A⎢−
⎥
ω
⎢⎣
0⎥
⎦
⇒
I=−
⇒
I=
A
⎡ cos ( ω t ) − cos 0° ⎤⎦
ω⎣
A
⎡ 1 − cos ( ω t ) ⎤⎦
ω⎣
At x = 0, v = v0 . Find the velocity v when the displacement becomes x.
Solution
The given equation is
dv
= −ω 2 x
dx
vdv = −ω 2 x dx
v
⇒
v
⇒
x
∫ vdv = −ω ∫ x dx
2
0
v0
v
Illustration 50
The velocity v and displacement x of a particle executing simple harmonic motion are related as
v
dv
= −ω 2 x
dx
01_Mathematical Physics_Part 1.indd 28
⇒
x
v2
x2
= −ω 2
2 v
2 0
0
⇒
v 2 − v02 = −ω 2 x 2
⇒
v = v02 − ω 2 x 2
11/28/2019 6:46:21 PM
Chapter 1: Mathematical Physics
1.29
Practice Exercise
Single Correct Choice Type Questions
This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
2.
If ax 2 + bx + c = 0 , a ≠ 0 , then
(A) x = −
b ± b 2 − 4 ac
2
(B)
x=−
b ± b 2 − 4 ac
4
(C) x = −
b ± b 2 − 4 ac
2a
(D) x = −
b ± b 2 − 4 ac
4a
4.
5.
6.
7.
8.
(B) x = 3 , y = 2
(D) x = −3 , y = −2
(B) 2
(D) 4
log e ( mn ) =
(A) log e m × log e n
(B)
log e m − log e n
⎛ m⎞
(C) log e ⎜ ⎟
⎝ n⎠
⎛ 1⎞
(D) log e m − log e ⎜ ⎟
⎝ n⎠
12. If 3 x 2 + 8 x + 5 = 0 , then
(A) log e ( x + y )
(B)
⎛ x⎞
(C) log e ⎜ ⎟
⎝ y⎠
⎛ y⎞
(D) log e ⎜ ⎟
⎝ x⎠
log e ( xy )
x=
(A) 2z = x + y
(B)
z = 2x + 2 y
(C) z = xy
(D) z = xy
log e x = k log10 ( x 2 ) , then k equals
(B) 4.606
(A) 2.303
(D) 3.303
(C) 1.151
If log e ( x 4 ) = k log10 x , then k equals
(A) 2.303
(B) 4.606
(C) 9.212
(D) 13.818
For x 1 , the value of ( 1 + x ) is
n
1
( 1 − nx )
2
(A) 1 − nx
(B)
(C) 1 + nx
1
( 1 + nx )
(D)
2
log a x equals
⎛ x⎞
(A) log e ⎜ ⎟
⎝ a⎠
(B)
log e x
log a e
log e x
log e a
(D)
log x e
log a e
log 2 ( 8 ) =
(A) 1
(C) 4
01_Mathematical Physics_Part 2.indd 29
13.
(B)
(C) x = −1
(D) x = −
5
3
log b a × log a b =
log a ( ab )
(A) 0
(B)
(C) 1
(D) log b ( ab )
14. In log a x , the value of a must be
(A) between 0 and 1
(B) positive, but not 1
(C) positive but not zero
(D) in some other interval
15. The ratio of area of circle of radius r and surface area
of sphere of radius r, is
(A)
1
4
(B) 4
(C)
3
4r
(D)
1
4r
16. An equation of straight line ay = bx + c is given, where
a , b and c are constants. The slope of the given
straight line is
(A) –
(B) 3
(D) 6
5
3
(A) x = 1
If log e x + log e y = 2 log e z , then
(C)
9.
11.
2 3 = 72 , then
log e x + log e y =
log10 100 =
(A) 1
(C) 3
x y
(A) x = 2 , y = 3
(C) x = −2 , y = −3
3.
10.
(C) b
a
b
(B)
b
a
(D) c
11/28/2019 6:44:13 PM
1.30 JEE Advanced Physics: Mechanics – I
(C)
17. Correct graph of 3 x + 4 y + 1 = 0 is
y
(A)
(B)
1
(D)
y
y
y
x
1/3
x
x
x
21. Graph of y = x + 2 is
(C)
(D)
y
y
(A)
18. Graph of y = 2x − 3 is
(C)
(B)
x
x
–1/3
y
y
x
x
–1/3
(A)
(B)
y
(D)
y
y
y
x
x
x
x
(C)
(D)
y
22. Graph of y = 2x − 3 is
y
(A)
x=3
19. Correct graph of y = 3 x + 4 is
y
x
(B)
y
y
x
x
20. Correct graph of y = x + 1 is
(B)
y
x
01_Mathematical Physics_Part 2.indd 30
x
x = –3/2
23. Graph of y = 2x + 1 + 1 is
y
(A)
(A)
x
y
(D)
x
x = –3
x
(D)
x = 3/2
y
(B)
y
x
y
(C)
x
(C)
y
x
x
(A)
(B)
y
(C)
y
x
x
(D)
y
x
y
x
11/28/2019 6:44:18 PM
1.31
Chapter 1: Mathematical Physics
(C)
24. Correct graph of y − 1 = x 2 is
(A)
(B)
y
(D)
y
y
y
x
x
x
x
28. Graph of y = 3 x 2 − 4 x + 1 is
(C)
y
(D)
(A)
y
x
(C)
2
(A)
(B)
y
y
x
x
25. Correct graph of y = − ( x + 2 ) is
(B)
y
x
(D)
y
y
y
x
x
x
x
29. Graph of x 2 y = 2 is best represented by
(C)
(D)
y
y
x
(A)
(B)
y
y
x
x
x
(C)
26. Correct graph of y = 2x 2 + 3 x + 1 is
(A)
y
(B)
(D)
y
y
y
x
x
x
x
30. Graph of y = 1 − e − x is best represented by ( for x > 0 )
(C)
(D)
y
y
(A)
(B)
y
y
I
x
x
x
x
2
27. Graph of y = 2 ( x − 1 ) + 2 is
(A)
(B)
y
(C)
–I
(D)
y
y
y
x
x
x
01_Mathematical Physics_Part 2.indd 31
x
11/28/2019 6:44:24 PM
1.32 JEE Advanced Physics: Mechanics – I
31. 1 radian is equal to
32.
π
degree
180
(B)
180
degree
π
(A)
2 tan θ
1 − tan 2 θ
(B)
2 tan θ
(C)
90
degree
π
(D)
18
degree
π
(C)
tan θ
1 − tan 2 θ
(D)
2 tan θ
1 + tan 2 θ
(A) cos ( 2θ )
(B)
sin ( 2θ )
(C) tan ( 2θ )
(D) cot ( 2θ )
⎛π⎞
tan ⎜ ⎟ =
⎝ 6⎠
(B)
1
2
(A) cot θ
(B)
(C) − cot θ
(D) tan θ
1
2
2 tan θ
1 + tan 2 θ
(B)
2sin θ
(C)
tan θ
1 + tan 2 θ
(D)
2 tan θ
1 − tan 2 θ
(A)
3 −1
2
(B)
3 +1
2
(C)
3 −1
2 2
(D)
3 +1
2 2
(A)
3 +1
2
(B)
3 +1
2 2
(C)
3 +1
4
(D)
3 −1
4
−1
44. The value of cos ( 75° ) is
sin ( 2θ ) =
(C)
(A)
43. The value of sin ( 75° ) is
(D) zero
(A) 2sin θ
1
sin θ cos θ
2
(B)
2sin θ cos θ
(D) sin θ cos θ
( 1 − cos θ ) equals
2⎛θ⎞
38.
− tan θ
⎛ 3π ⎞
cos ⎜
=
⎝ 2 ⎟⎠
(C)
sin ( 2θ ) equals
42. The value of sin ( 15° ) is
1
2
⎛ 3π
⎞
tan ⎜
+θ⎟ =
⎝ 2
⎠
(B)
cos 2 θ − sin 2 θ equals
−1
(D) −
(A) 1
37.
41.
1
3
⎛ 3π ⎞
tan ⎜
=
⎝ 4 ⎟⎠
(C)
36.
1
3
(D) −
(A) 1
35.
40.
(B)
3
(C) − 3
34.
tan ( 2θ ) equals
(A)
(A)
33.
39.
2
2 cos θ
(A) 2 cos ⎜ ⎟
⎝ 2⎠
(B)
(C) 2 sin 2 θ
⎛θ⎞
(D) 2 sin 2 ⎜ ⎟
⎝ 2⎠
( 1 + cos θ ) equals
2⎛θ⎞
2
2 cos θ
(A) 2 cos ⎜ ⎟
⎝ 2⎠
(B)
(C) 2 sin 2 θ
⎛θ⎞
(D) 2 sin 2 ⎜ ⎟
⎝ 2⎠
01_Mathematical Physics_Part 2.indd 32
45.
(A)
3 −1
2
(B)
3 +1
2
(C)
3 −1
2 2
(D)
3 +1
2 2
( cos A − cos B ) =
⎛ A+B⎞
⎛ A −B⎞
(A) 2 cos ⎜
sin ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
(B)
⎛ A+B⎞
⎛ A −B⎞
2 sin ⎜
cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎛ A+B⎞
⎛ B− A⎞
(C) 2 sin ⎜
cos ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
⎛ A+B⎞
⎛ B− A⎞
(D) 2 cos ⎜
sin ⎜
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
11/28/2019 6:44:33 PM
Chapter 1: Mathematical Physics
46.
sin 2 ( 37° ) + sin 2 ( 53° ) =
(A) 0
47.
(B) 1
1
2
(C)
55.
(B)
sin θ can never have a value
(A) 1
(B)
1
4
(C)
−1
(D) 2
50.
51.
(A) −1 and 1
(B)
(C) zero and 1
(D) zero and 2
sin ( 100π ) is equal to
(A) 1
(B) 100
(C) zero
(D)
56.
(A) − cos θ
(B)
cos θ
(C) sin θ
(D) − sin θ
57.
x2 − 9
=
x→3 x − 3
58.
x 2 − 25
lim
=
x →−5 x + 5
−10
(A) 10
(B)
(C) 0
(D) 5
59.
( )
5
5 2
x
2
(C)
5 2
x
2
7
54.
(A)
1 ⎛ du u dv ⎞
−
⎜
⎟
v ⎝ dx v dx ⎠
(B)
1 ⎛ v du dv ⎞
− ⎟
⎜
u ⎝ u dx dx ⎠
(C)
1 ⎛ u du dv ⎞
− ⎟
⎜
u ⎝ v dx dx ⎠
(D)
1 ⎛ u dv du ⎞
−
⎜
⎟
v ⎝ v dx dx ⎠
d ( n)
u =
dx
(B)
nun−1
(C) zero
du
dx
(D) None of these
d
( sin u ) =
dx
(A) cos u
(B)
− cos u
du
dx
3
(B)
5 2
x
2
60.
7
2 2
(D)
x
7
d
(u + v − w) =
dx
(B) 1 + tan 2 x
(D) None of these
(A) − sec x tan x
(B)
sec 2 x cot x
sin x
cos 2 x
(D)
cos x
sin 2 x
d
( cosec x ) =
dx
⎛ du dv ⎞ dw
(A) ⎜
±
⎝ dx dx ⎟⎠ dx
(B)
−
du dv dw
+
−
dx dx dx
(A) − cosec x tan x
(B)
−
du dw dv
−
+
dx dx dx
(D)
du dv dw
−
+
dx dx dx
(C) −
cos x
sin 2 x
(D)
cos x
sin 2 x
(C)
01_Mathematical Physics_Part 2.indd 33
du
dx
d
( sec x ) =
dx
(C)
61.
(D) − ( cos u )
d
( tan x ) =
dx
(A) sec x tan x
(C) sec x
d
x2 =
dx
(A)
d ⎛ u⎞
⎜ ⎟=
dx ⎝ v ⎠
(C) ( cos u )
5
53.
du
dz
dw
dv
− uvz
+ uwz
+ vwz
dx
dx
dx
dx
(A) nun−1
lim
(B) 4
(D) 6
dz
dw
dv
du
− uvz
− uwz
− vwz
dx
dx
dx
dx
dz
dv
du
dw
+ vwz
+ uvz
+ uvw
dx
dx
dx
dx
(D) − uvw
1
2
cos ( 180 − θ ) is equal to
(A) 3
(C) 5
52.
−1 and zero
du
dz
dw
dv
+ uvz
− uwz
− vwz
dx
dx
dx
dx
− uvw
(C) uwz
48. The value of sin 2 θ always lies between
49.
d
( uvwz ) =
dx
(A) uvw
4
(D)
5
1.33
sin x
cos 2 x
11/28/2019 6:44:42 PM
1.34 JEE Advanced Physics: Mechanics – I
62.
63.
64.
65.
d ( 5
x + x7 + x 9 ) =
dx
71.
(A)
x 6 x 8 x10
+
+
6
8
10
(B)
(C)
x 5 x7 x 9
+
+
5
7
9
(D) 5x 3 + 7 x 5 + 9x 7
5x 4 + 7 x 6 + 9x 8
72.
d
( log e x ) =
dx
(A)
1
e
(B)
1
log e x
(C)
1
x
(D) x log e x − x
d ( 2x )
e
=
dx
(B)
(C) 2e 2 x
(D) 2e x
ex
d
( x log e x − x ) =
dx
(A) zero
(B) 1
(C) log e x
(D) x log e x
75.
66. The tangent to the curve y 2 = 4 x at ( 1, 2 ) is inclined
to the x-axis at an angle of
(A)
π
6
(B)
π
3
(C)
π
2
(D)
π
4
67. The maximum value of the function y = sin x + cos x is
(A) 1
(B) 2
(C)
68.
69.
70.
2
(D) − 2
76.
77.
d
⎡ log e ( sin x ) ⎤⎦ =
dx ⎣
− cot x
(A) cot x
(B)
(C) tan x
(D) − tan x
d
⎡ log e ( cos x ) ⎤⎦ =
dx ⎣
d ( 2
x sin x ) =
dx
(A) x 2 cos x
(B)
(C) x 2 cos x + 2x cos x
(D) x 2 cos x + 2x sin x
78.
(A) tan θ = 7
(B)
(C) tan θ = 11
(D) tan θ = 12
(A)
1
+ sec x tan x
x
(B)
x + sec x tan x
(C)
1
+ sec 2 x
x
(D)
1
− sec 2 x
x
(A) e x cos x
(B)
e x ( cos x + sin x )
(C) −e x cos x
(D) e x ( sin x − cos x )
d ( x
e sin x ) =
dx
d
( x log e x ) =
dx
(A) 1
(B) 1 + log e x
(C) 1 − log e x
(D) log e x
d ( x
e + 6x ) =
dx
(A) e x +6 log e 6
(B)
(C) e x log e 6
(D) e x + 6 x log e 6
6 x + e log 6 e
d ⎛1
⎞
2
⎜⎝ + tan x + x + log e x ⎟⎠ =
dx x
1
1
+ sec 2 x + 2x +
2
x
x
1
1
2
− sec x − 2x +
x
x2
(A) −
− cot x
(B)
(C) tan x
(D) − tan x
(B)
(B) − sec x
(D) − cosec x
1
1
− sec 2 x + 2x −
x
x2
1
1
(D) − 2 + sec x tan x + 2x +
x
x
01_Mathematical Physics_Part 2.indd 34
tan θ = 10
d
( log e x + tan x ) =
dx
(A) cot x
d
⎡ log e ( sec x + tan x ) ⎤⎦ =
dx ⎣
(A) sec x
(C) cosec x
x 2 sin x + x 2 cos x
73. The tangent to the curve y = 3 x 2 − 5 at the point
( 2, 7 ) makes an angle q with the positive x-axis. Then
74.
(A) e 2 x
d
⎡ log e ( cosec x + cot x ) ⎤⎦ =
dx ⎣
(B) sec x
(A) − sec x
(C) − cosec x
(D) cosec x
(C) −
11/28/2019 6:44:54 PM
Chapter 1: Mathematical Physics
79.
d ⎛ tan x + cot x ⎞
⎜
⎟⎠ =
x
dx ⎝
85.
(A)
x
dy
=1
dx
d2 y
(C)
⎛ sec 2 x − cosec2 x ⎞ ⎛ tan x + cot x ⎞
⎜⎝
⎟⎠ − ⎜⎝
⎟⎠
x
x2
(B)
y = x log e x
(A)
( sec2 x − cosec2 x )
86.
⎛ tan x + cot x ⎞
(C) − ⎜
⎟⎠
⎝
x2
dx
81.
d ( x
e tan x ) =
dx
(B)
(C) e x ( tan x + sec 2 x )
(D) e x ( tan x − sec 2 x )
1
2 x
(C) 10 x 4 − 8 x +
1
2x x
(D) 10 x 4 − 8 x −
88.
1
2 x
d ⎛ 3x 2 + 1 ⎞
⎜
⎟=
dx ⎝ 2x − 1 ⎠
89.
d2 y
=x
dx 2
(A)
dy
= e x ( cos x − sin x )
dx
(B)
dy
= e x cos x
dx
(C)
d2 y
= 2e x sin x
dx 2
(D)
d2 y
= 2e x cos x
dx 2
(B)
1
−1 2
sec 2 x ( tan x )
2
d
tan x
dx
−1 2
1
( tan x )−1 2
2
(D) 2 ( tan x )
−1 2
d
sin ( log x )
dx
(A) cos ( log x )
(B)
(C) x cos ( log x )
(D)
d
dx
log ( cos x )
cos ( log x )
x
( 2x 2 + 1 )
6x
2x − 1
(A) 2x ( 2x 2 + 1 )
(B)
2x ( 2x 2 + 1 )
(B)
( 2x − 1 ) ( 6 x + 1 ) − ( 3 x 2 + 1 ) 2x
( 2 x − 1 )2
(C)
( 2x 2 + 1 )1 2
(D)
( 2x 2 + 1 )−1 2
(B)
2xe 2 x
12
3
2
90.
12x − 6 x − 8 x
4x 2 − 4x + 1
(
)
ex
x
1⎞
(A) e ⎜ log e x + ⎟
⎝
x⎠
(B)
(C) e x log e x + 1
(D) None of these
x⎛
e 2x
2x
(
(C) e 2 x
d x
e log e x =
dx
91.
(A)
dy
= − cos x + 4 x 3
dx
(B)
dy
= sin x + 4 x 3
dx
(C)
d2 y
= − sin x + 12x 2
dx 2
(D)
d2 y
= − cos x + 6 x 2
dx 2
(D) e 2 x
)−1 2
d ( 4
x − 2 sin x + 3 cos x )
dx
(A) 4 x 3 − 2 cos x + 3 sin x
(C) 4 x 3 + 2 cos x − 3 sin x
d 2
x sin x log x
92.
dx
(
y = sin x + x 4
−1 2
d ( 2x )
e
dx
(A)
(D) None of these
84.
(D)
(A)
(C)
83.
dy
= log e x
dx
y = e sin x
(C)
(B) 10 x 4 − 8 x − x
1
x
(A) 2 sec 2 x ( tan x )
d ⎛ 5
1 ⎞
2
⎜⎝ 2x − 4 x −
⎟=
dx
x⎠
(A) 2x 5 − 8 x −
82.
87.
e x sec 2 x
(A) e x sec x tan x
=
2
(B)
x
(D) None of these
80.
1.35
(B) 3 x 2 + 2 cos x + 3 sin x
(D) 4 x 3 − 2 cos x − 3 sin x
)
(A) 2x sin x log x + x 2 cos x log x + x sin x
(B)
x 2 sin x log x + 2x cos x log x + x sin x
(C) 2x sin x log x + x 2 cos x log x + sin x
(D) None of these
01_Mathematical Physics_Part 2.indd 35
11/28/2019 6:45:03 PM
1.36 JEE Advanced Physics: Mechanics – I
93.
d ( x2 + 1 )
dx x + 1
(A)
(C)
94.
x 2 + 2x − 1
(B)
( x + 1 )2
2
x + 2x − 1
x+1
xy = c 2 , then
(A)
100. If y = x 3 + 2x + 1, then
(D)
( x + 1 )2
2
x + 2x + 1
101.
( x + 1 )2
dy
dx
y=
dy
1+ x
is equal to
, then
x
dx
e
(A)
x
ex
(B)
x
y
y
x
(D) −
95. If x = at 2 and y = 2 at , then
102.
y
x
1
t
(C) 1
(D) None of these
(B)
1
(D)
( x + 2 )3
dy
π
at x =
is
dx
3
(A) 6
(C) 8
104.
dy
1+ x
is equal to
then
x
dx
e
x
ex
9 − 96 3
4
(B)
9 − 86 3
4
(A)
(C)
9 − 76 3
2
(D) None of these
(C)
dy
is
dx
105.
(A) 4 x cos ( 2x 2 )
(B)
2 cos ( 2x 2 )
(C) 4 cos ( 2x 2 )
(D) −4 cos ( 2x 2 )
98. The minimum value of y = 2x 2 − x + 1 is
(B)
( x + 1)
−1
( x + 2 )3
x
− x
e
(D) None of these
ex
∫ log x =
e
1
x
(A)
(B)
(C) x log e x − x
106.
−x
( x + 2 )3
dy
at x = 1 is
dx
(B) 7
(D) 5
y=
(A)
97. If y = sin ( 2x 2 ) , then
x
( x + 2 )3
103. If y = x 3 + 2x + 1 then
(B)
96. If y = sin 3 x − 3 sec 2 x , then
d ⎛ x+1 ⎞
dx ⎜⎝ ( x + 2 )2 ⎟⎠
(C)
(A) t
x
ex
(D) None of these
ex
(A)
dy
dx
(B) –
( x + 1)
(C)
x
y
(C) −
(A) 6
(C) 8
x 2 − 2x + 1
dy
at x = 1 is
dx
(B) 7
(D) 5
1
e
(D) x log e x
∫ x dx for n = −1 is
n
(A) −
3
8
(B)
−
5
8
(A) Not defined
(B)
(C) −
7
8
(D) −
9
8
(C) log e x
(D) 2log e x
dy
π
at x =
99. If y = sin x − 2 tan x , then
is
4
dx
2
2
(A) −11
(B)
(C) −13
(D) −15
01_Mathematical Physics_Part 2.indd 36
−7
107.
∫ ( x + x + x ) dx =
5
7
x n+ 1
n+2
9
(A) 5x 4 + 7 x 6 + 9x 8
(B)
x 5 x7 x 9
+
+
5
7
9
⎛
x3 x5 ⎞
(C) x 5 ⎜ x +
+ ⎟
⎝
3
5 ⎠
(D)
x 6 x 8 x10
+
+
6
8
10
11/28/2019 6:45:12 PM
Chapter 1: Mathematical Physics
b
108.
∫
2
a
π 2
dx
=
x
115. The value of
2log e ( b − a )
(B)
⎛ b2 ⎞
(C) log e ⎜ 2 ⎟
⎝a ⎠
⎛ a⎞
(D) 2log e ⎜ ⎟
⎝ b⎠
5
116. Value of
dx
∫ 2 + 3x =
111.
112.
1
log e ( 2 + 3 x )
3
(B)
1
log ( 3 + 2x )
3
(C)
1
⎛ 2 + 3x ⎞
log e ⎜
⎝ 2 ⎟⎠
3
(D)
1
⎛ 2 ⎞
log e ⎜
⎝ 2 + 3 x ⎟⎠
3
⎛ 13 ⎞
(A) ln ⎜ ⎟
⎝ 9 ⎠
(C) 20 x 3
(D) x 5
1
− cos ( 2x )
2
(B)
(C) − cos x
(D) None of these
118.
(D) None of these
1
∫ (3 − 2x) dx is
2
1
9
(B)
(C) −
4
9
(D) None of these
( 1 + cos x )3 2
3
2
3
x
1
+
− 2x
3 x2
(B)
(D)
3
x
1
−
+ 2x
3 x2
(B)
(C) log 1 + tan x + C
(D) − log 1 + tan x + C
119.
x
2 2 cos + C
2
(D) None of these
∫ 2t dt is equal to
(A) 0
(B) 4
(C) 2
(D)
1
2
1
2
π 2
120.
∫ sin x dx is equal to
π 6
∫ ( 3x − 4x + 1 ) dx is
2
0
01_Mathematical Physics_Part 2.indd 37
(B)
0
log 1 + cot x + C
1
(A) 0
(C) 2
2
9
2
cosec 2 x dx
=
1 + cot x
(A) − log 1 + cot x + C
+C
⎛ x⎞
(C) 2 2 sin ⎜ ⎟ + C
⎝ 2⎠
x3 1
− + 2x
3 x
−
∫ ( 1 + cos x ) dx =
2
114. Value of
1 ⎛ 13 ⎞
ln ⎜ ⎟
2 ⎝ 9 ⎠
(A) −
(A)
1⎞
⎛
⎜⎝ x + ⎟⎠ dx =
x
1
(A) x + + 2x
x
∫
1
117. Value of
6x 2
(A) − cos ( 2x )
3
113.
1 ⎛ 15 ⎞
ln ⎜ ⎟
2 ⎝ 9⎠
sin ( 2x ) dx =
(C)
(B)
0
(B)
∫
(C)
5x 4 dx =
(A) 4 x 3
∫
1
∫ 2x + 3dx is
3
(A)
∫
(B) 1
(D) 2
(A) 0
(C) −1
0
110.
∫ sin ( 2θ ) dθ is
0
(A) log e b − log e a
x
109.
1.37
(B) 1
(D) 3
(A)
1
2
(B)
(C)
3
2
(D) 0
11/28/2019 6:45:20 PM
1.38 JEE Advanced Physics: Mechanics – I
121.
dt
∫ ( 6t − 1 ) is equal to
(A)
1
log e ( 6t − 1 ) + C
6
1
(C) − log e ( 6t − 1 ) + C
6
01_Mathematical Physics_Part 2.indd 38
122.
(B)
log e ( 6t − 1 ) + C
(D) None of these
∫ ( 4 cost + t ) dt is equal to
2
t3
+C
3
(B)
t3
+C
3
(D) 4 sin t + 2t 3 + C
(A) −4 sin t +
(C) 4 sin t +
−4 sin t + t 2 + C
11/28/2019 6:45:21 PM
1.39
Chapter 1: Mathematical Physics
Answer Key—Practice Exercise
Single Correct Choice Type Questions
1. C
2. B
3. B
4. D
5. C
6. C
7. C
8. C
9. B
10. B
11. D
12. C, D
13. C
14. B
15. A
16. B
17. D
18. D
19. B
20. B
21. C
22. B
23. A
24. A
25. D
26. A
27. B
28. C
29. C
30. A
31. B
32. B
33. B
34. C
35. D
36. B
37. D
38. A
39. A
40. A
41. A
42. C
43. B
44. C
45. B, C
46. B
47. D
48. C
49. C
50. A
51. D
52. B
53. B
54. C
55. C
56. A
57. B
58. C
59. B
60. C
61. C
62. B
63. C
64. C
65. C
66. D
67. C
68. A
69. D
70. A
71. C
72. D
73. D
74. C
75. B
76. B
77. D
78. A
79. B
80. C
81. C
82. C
83. A
84. C
85. C
86. D
87. B
88. D
89. B
90. A
91. D
92. A
93. A
94. D
95. B
96. A
97. A
98. C
99. D
100. D
101. B
102. B
103. D
104. B
105. C
106. C
107. D
108. C
109. C
110. D
111. B
112. B
113. A
114. A
115. B
116. B
117. B
118. C
119. B
120. C
121. A
122. C
01_Mathematical Physics_Part 2.indd 39
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01_Mathematical Physics_Part 2.indd 40
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CHAPTER
2
Measurements and
General Physics
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Physical Quantity and its
(d) Principle of Homogeneity (g) Errors and their Propagation
Measurement
and its uses
(h) Measurements done using
(b) Fundamental and Derived (e) Limitations of Dimensional
Vernier Calliper (VC)
Units
Analysis
(i) Measurements done using
(c) Dimensional Analysis
(f) Least Count, Significant
Screw Gauge (SG).
Figures and Rounding off
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
and Advanced) are also given.
scientific Process
Objective Observation
The history of science reveals that it has evolved
through a series of steps. Let us have a small discussion of the steps involved.
An observation that remains identical for all
the observers (persons) is called an Objective
Observation.
STEP-1: Observation
STEP-2: Proposing/propounding a theory based on
those observations.
STEP-3: Verification of the theory as applied to those
observations.
STEP-4: Modification in the theory, if at all necessary.
EXAMPLE:
Four observers viewing the same painting may feel differently about the “beauty” of the painting, where as
they shall report identical length, breadth or area of the
painting, when asked. So, Beauty is not an objective
observation as it cannot be assigned a numerical
value along with some appropriate unit (that could have
measured it).
Physics always deals with objective observation.
obserVation
Observations are basically of two types
Subjective Observation
Physical Quantity
An observation that varies from person to person is
called Subjective Observation. Physics never deals
with subjective observations like beauty, emotion,
personality etc.
The objective quantities to which a numerical value
can be attached along with some unit (specified to
measure it) are called Physical Quantities. Else, they
may also be defined as the quantities through which
02_Measurements, General Physics_Part 1.indd 1
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2.2 JEE Advanced Physics: Mechanics – I
(through the help of which) we can describe the Laws
of Physics appropriately. A physical quantity can be
completely specified if it has
(a) only magnitude (some constants or some ratios)
e.g., specific gravity, dielectric constant, strain,
refractive index etc.
(b) magnitude along with a unit (scalars) e.g., mass,
length, time, energy, current etc.
(c) magnitude along with a unit in a specified direction (vectors) such that laws of vector algebra are
obeyed by this quantity e.g., displacement, force,
momentum, torque, angular momentum, electric
field, magnetic field etc.
(d) magnitude and phase (phasors) e.g., superposition of mechanical wave, AC voltages and currents, SHM etc.
(e) magnitude of varying values in different directions (having no specified d
­ irections) (are called
Tensors) e.g., Moment of Inertia (cannot be
defined without specifying the axis of rotation).
So, measurement of a Physical Quantity = nu
where, u is the unit selected (of same nature) to
­measure the physical quantity.
n is the number of times this selected unit is contained in it.
For a particular measurement, irrespective of the system in which the unit is selected, measurement of a
physical quantity is a constant.
⇒
nu = constant
⇒
n1u1 = n2 u2 = n3 u3 = ................. where,
u1 , u2 , u3 , ..…. are the units selected to measure a
physical quantity in system 1, 2, 3, …….. respectively.
n1 , n2 , n3 , …… are the numerical values that the
measured physical quantity contains corresponding
to the respective systems.
Also, we observe that
In few anisotropic media the physical quantity(ies)
like density, refractive index, dielectric constant,
stress, strain, electric conductivity all become Tensors.
1
u
i.e., the bigger the unit selected to measure the
physical quantity, the smaller the numerical values
vice-versa.
Measurement of a Physical Quantity
FUNDAMENTAL AND DERIVED UNITS
The process of measurement is basically a comparison process. To measure a physical quantity we need
to have the following two things.
The exact specification of the measurement of a
physical quantity requires
(a) Firstly, a unit u (of same nature) to measure that
physical quantity.
(b) Secondly, the number of times ( n ) this selected
unit is contained in the required physical
quantity.
e.g., if we are asked to measure the length and breadth
of a room with the help of a scale that is half a metre
in length (half metre scale) and we observe that this
unit (half metre scale) is contained 10 times in the
length of the room and 8 times in its width, then
⎛1 ⎞
Length of the room = 10 ⎜ m ⎟ = 5 m
⎝2 ⎠
⎛1 ⎞
Breadth of the room = 8 ⎜ m ⎟ = 4 m
⎝2 ⎠
02_Measurements, General Physics_Part 1.indd 2
n∝
(a) the standard or unit in which the quantity is
measured and
(b) the numerical value representing the number of
times the quantity contains that unit.
The physical quantities which do not depend upon
other quantities are called fundamental ­quantities.
In M.K.S. system the fundamental quantities are
mass, length and time, while in more general
Standard International (S.I.) system the fundamental quantities are mass, length, time, temperature,
illuminating power (or luminous intensity), current
and amount of substance. The units of fundamental
quantities are called fundamental units. The units
of physical quantities which may be derived from
fundamental units are called derived units.
11/28/2019 6:48:59 PM
Chapter 2: Measurements and General Physics
Conceptual Note(s)
Measurement of a Physical Quantity means a comparison of it with some reference standard, also called
as unit, which does not change (under any circumstances). So, these standards must be
(a) invariable
(b) easily accessible
(c) precise and
(d) universally agreed.
Fundamental Quantity
Unit
Symbol
Time
second
s
Temperature
kelvin
K
Luminous Intensity
candela
cd
Electric Current
ampere
A
Amount of Substance
mole
mol
2.3
In S.I. system there are two supplementary units.
(a) Radian (rad): Unit of plane angle.
(b) Steradian (sr): Unit of solid angle.
System of Units
Following principal systems of units are used
commonly.
C.G.S. System
In this system the unit of length is centimetre (cm),
that of mass is gram (g) and that of time is second (s).
F.P.S. System
In this system the unit of length is foot, weight is
pound (lb) and time is second.
Conceptual Note(s)
(a) In fps system the unit pound is the unit of weight
and not of mass.
(b) 1pound = 0.4536 kgwt
i.e. 1 pound is equivalent to the weight of a body
having a mass of 0.4536 kg.
(c) The unit of mass in fps system is slug.
Conceptual Note(s)
To understand the concept of solid angle, let us consider a football having black and white patches on it.
Now consider any one patch, say black. Then on the
boundary of this patch lie infinite number of points.
If you join all these points with the centre of the football, then you observe all these points to be at equal
distance from it and this distance happens to be the
radius of the football. Now, if you join all these points
to the centre of the football then the angle enclosed
by this patch at the centre of the football is called the
Solid Angle, just like the angle at the end of an empty
ice-cream cone. Now this solid angle, denoted by Ω ,
has a value given by
Ω=
Area of patch on the surface of sphere
( Radius of sphere )2
M.K.S. System
In this system the units of length is metre (m), mass is
kilogram (kg) and time is second(s).
S.I. System
Patch on
surface
of sphere
R
Ω
R
R
In this system there are seven fundamental quantities
whose units and symbols are as follows:
Fundamental Quantity
Unit
Symbol
Length
metre
m
Mass
kilogram
kg
(Continued)
02_Measurements, General Physics_Part 1.indd 3
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2.4 JEE Advanced Physics: Mechanics – I
Prefixes for Power of Ten
Prefix
Abbreviation
Power of Ten
Prefix
Abbreviation
Power of Ten
mega
M
106
atto
a
10-18
giga
G
109
femto
f
10-15
tera
T
1012
pico
p
10-12
peta
P
1015
nano
n
10-9
exa
E
1018
micro
m
10-6
milli
m
10-3
centi
c
10-2
kilo
k
103
EXAMPLES:
Derived units
(Continued)
−6
1 micro second = 1 μs = 10 s
−9
1 nano second = 1 ns = 10 s
−3
1 kilo-metre = 1 km = 10 m
Physical quantity
Dimensional physical quantity
Dimensionless physical quantity
Dimensional constant
Dimensional variable
Dimensionless constant
Dimensionless variable
e.g. Plank’s Constant (h),
Universal Gravitational
Constant (G), Boltzmann
Constant (kB), Universal
Gas Constant (R), Stefan’s
Constant (σ ) etc.
e.g. Length, mass, time,
energy, momentum,
torque, resistance, charge,
current magneticfield,
electric field etc.
e.g. π , 0, 1, 2, 3, ....
Fine Structure Constant (α )
2
α= e
= 1
2hε oC 137
e.g. Strain, plane angle,
solid angle, Mach number,
Renolds number
Some Important Commonly Used Units
(a) micrometer ( μm ) = 1μm = 10 −6 m
(b) Angstrom 1 Å = 10 −10 m
(c) Astronomical unit
This is the mean distance of earth from sun
11
11
1 AU = 1⋅ 496 × 10 m ≈ 1⋅ 5 × 10 m
(d) Light year
It is the distance traversed by light in vacuum in
1 year,
1light year = 365 × 24 × 60 × 60 × 3 × 108 = 9 ⋅ 45 × 1015 m
02_Measurements, General Physics_Part 1.indd 4
(e) parsec
It is the distance at which an arc of length one
Astronomical unit subtends an angle of 1 second
i.e. 1″.
Since l = rq
l
⇒ r=
θ
⎛ 1 ⎞⎛ π ⎞
where l = 1 AU and θ = 1′′ = ⎜
rad
⎝ 3600 ⎟⎠ ⎜⎝ 180 ⎟⎠
⇒ r = 1 parsec =
1.496 × 1011 m
1
π
×
(rad)
3600 180
11/28/2019 6:49:01 PM
Chapter 2: Measurements and General Physics
⇒ r = 1 parsec = 3 ⋅ 07 × 1016 m
(l) 1 ft = 0 ⋅ 3048 m
⇒ 1 parsec = 3 ⋅ 26 light years
(m) 1 pound = 453.6 g = 0 ⋅ 4536 kg
(f) X-ray Unit ( XU ) = 1 XU = 10 −13 m
(n) 1 slug = 14.57 kg
(g) 1 Bar = 105 Nm−2 = 105 pascal
(o) 1 poiseuille ( Pl ) = 10 poise
(h) 1 atmosphere ( atm ) = 1.013 × 105 Pa
(p) 1 metricton = 10 quintal = 1000 kg
2.5
(q) 1 Chandra Shekhar Limit (CSL) = 1⋅ 4 Ms
(i) 1 torr = 1 mm of Hg = 133.3 Pa
where Ms = mass of sun
(j) 1 barn = 10 −28 m2
(r) 1 Shake = 10 −8 s
(k) 1 horse power = 746 watt
DIMENSIONS
The powers to which the fundamental units of mass,
length and time are raised so as to get the required
physical quantity.
EXAMPLE:
To get the physical quantity “force”, mass has to be raised
to the power 1, length to the power 1 and time to the
power -2. So, dimensions of force are 1 in mass, 1 in
length and -2 in time.
f­ormula of force is MLT −2 . Whenever a physical
quantity is written in square brackets, it just means
that dimensional formula of the physical quantity
has to be taken.
Dimensional Equation
Whenever a physical quantity is equated to its dimensional formula, we get a dimensional equation. So,
dimensional equation for force is
Dimensional Formula
−2
[ F ] = MLT
M a Lb T c is the dimensional formula of a physical quantity which has dimensions a, b and c in
mass, length and time respectively. So, dimensional
a b c
[ X ] = M L T
In general, any physical quantity X, having dimensional formula M a Lb T c , has a dimensional equation
DIMENSIONS OF SOME PHYSICAL QUANTITIES
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
1.
Area ( A )
Length × Breadth
m2
M 0 L2T 0
2.
Volume ( V )
Length × Breadth × Height
m3
M 0 L3T 0
(Continued)
02_Measurements, General Physics_Part 1.indd 5
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2.6 JEE Advanced Physics: Mechanics – I
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
3.
Mass density ( ρ )
Mass
Volume
kgm −3
ML−3T 0
4.
Surface Mass
Density ( σ )
Mass
Area
kgm −2
ML−2T 0
5.
Linear Mass Density
(λ )
Mass
Length
kgm −1
ML−1T 0
6.
Frequency ( ν )
1
Time period
Hz
M 0 L0T −1
7.
Velocity, Speed ( v )
Displacement Distance
,
Time
Time
ms −1
M 0 LT −1
8.
Acceleration ( a )
Velocity
Time
ms −2
M 0 LT −2
9.
Force ( F )
Mass × Acceleration
newton (N)
MLT −2
10.
Impulse ( I )
Force × Time
Ns
MLT −1
11.
Work, Energy (W, E)
Force × Distance
joule (J)
ML2T −2
12.
Power ( P )
Work
Time
watt (W)
ML2T −3
13.
Momentum ( p )
Mass × Velocity
kgms −1
MLT −1
14.
Kinetic energy
( K or K.E. )
⎛ 1⎞
2
⎜⎝ ⎟⎠ × Mass × ( Velocity )
2
J
ML2T −2
15.
Potential energy (U)
Mass × Acceleration due to gravity × Height
J
ML2T −2
16.
Spring Constant ( k )
Force
Extension
Nm −1
M L0 T −2
17.
Elastic Potential
Energy ( U )
1
( Spring constant ) ( Extension )2
2
J
M L2 T −2
18.
Angle, Angular
displacement ( θ )
Arc
Radius
radian
M 0 L0T 0
(Continued)
02_Measurements, General Physics_Part 1.indd 6
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Chapter 2: Measurements and General Physics
2.7
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
19.
Trigonometric ratio
( sin θ , cosθ , tan θ ,
etc.)
Length
Length
No Units
M 0 L0T 0
20.
Angular velocity
(ω )
Angle
Time
rads −1
M 0 L0T −1
21.
Angular acceleration
(α )
Angular velocity
Time
rads −2
M 0 L0T −2
22.
Radius of gyration
(k )
Distance
m
M 0 LT 0
23.
Moment of inertia
(I )
Mass × ( Radius of gyration )
kgm 2
ML2T 0
24.
Angular momentum
(L)
Moment of inertia × Angular velocity
kgm 2s −1
ML2T −1
25.
Moment of force,
moment of couple
(τ )
Force × Distance
Nm
ML2T −2
Angular momentum
or Force × Distance
Time
Nm
ML2T −2
Sl.
2
26.
Torque ( τ )
27.
Angular frequency
(ω )
2π × Frequency
rads −1
M 0 L0T −1
28.
Rotational kinetic
energy ( R KE )
1
2
× Moment of inertia × ( Angular velocity )
2
J
ML2T −2
29.
Angular impulse
(J)
Torque × Time
Js
ML2T −2
30.
Centripetal
acceleration ( ac )
( Velocity )2
ms −2
M 0 LT −2
Force
Area
Nm −2
ML−1T −2
Radius
31.
Pressure ( P )
32.
Stress
Restoring force
Area
Nm −2
ML−1T −2
33.
Strain
Change in dimension
Original dimension
No units
M 0 L0T 0
34.
Modulus of elasticity
(E)
Stress
Strain
Nm −2
ML−1T −2
(Continued)
02_Measurements, General Physics_Part 1.indd 7
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2.8 JEE Advanced Physics: Mechanics – I
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
35.
Surface tension
( T or σ )
Force
Length
Nm −1
ML0T −2
36.
Surface energy
Energy
Area
Jm −2
ML0T −2
37.
Speed gradient
Speed
Distance
s −1
M 0 L0T −1
38.
Pressure gradient
Pressure
Distance
Nm −3
ML−2T −2
39.
Pressure energy
Pressure × Volume
Nm ( = J )
ML2T −2
40.
Fluid flow rate ( V )
m 3s −1
M 0 L3T −1
41.
Bulk modulus
( B ) or
( Compressibility )−1
Nm −2
ML−1T −2
( = Nm )
−1
4
( Pressure ) × ( Radius )
⎛π⎞
⎜⎝ ⎟⎠
8 ( Viscosity coefficient ) × ( Length )
Volume × ( Change in pressure )
( Change in volume )
42.
Coefficient of
viscosity ( η )
Force
Area × Speed gradient
kgm −1s −1
ML−1T −1
43.
Wavelength ( λ )
Distance
m
M 0 LT 0
44.
Hubble constant
(H)
Recession speed
Distance
s −1
M 0 L0T −1
45.
Solid Angle ( Ω )
steradian
M 0 L0T 0
Wm −2
ML0T −3
Area of patch on surface of sphere
( Radius of sphere )2
46.
Intensity of wave ( I )
⎛ Energy ⎞
⎜⎝
⎟
Time ⎠
Area
47.
Radiation pressure
(P)
Intensity of wave
Speed of light
Wm −3s
ML−1T −2
48.
Energy density ( u )
Energy
Volume
Jm −3
ML−1T −2
(Continued)
02_Measurements, General Physics_Part 1.indd 8
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Chapter 2: Measurements and General Physics
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
49.
Critical velocity
( vc )
Renold’s number × Coefficient of viscocity
Mass density × Diameter
ms −1
M 0 LT −1
50.
Escape velocity ( ve )
2 × Acceleration due to gravity × Earth’s radius
ms −1
M 0 LT −1
51.
Heat energy, internal
energy ( Q , U )
Work ( = Force × Distance )
J
ML2T −2
52.
Efficiency ( η )
Output work or energy
Input work or energy
53.
Gravitational
constant ( G )
Force × ( Distance )
Mass × Mass
54.
Planck constant ( h )
55.
Heat capacity (C),
entropy ( S )
56.
2
M 0 L0T 0
Nm 2kg −2
M −1L3T −2
Energy
Frequency
Js
ML2T −1
Heat energy
Temperature
JK −1
ML2T −2 K −1
Specific heat
capacity ( c )
Heat Energy
Mass × Temperature
Jkg −1K −1
M 0 L2T −2 K −1
57.
Latent heat ( L )
Heat energy
Mass
Jkg −1
M 0 L2T −2
58.
Thermal expansion
coefficient or
Thermal expansivity
( α , β or γ )
Change in dimension
Original dimension × temperature
K −1
M 0 L0 K −1
Heat energy × Thickness
Area × Temperature × Time
Js −1m −1K −1
( = Wm −1K −1 )
MLT −3 K −1
⎛ Energy ⎞
⎜⎝
⎟
Area × Time ⎠
Wm −2K −4
ML0T −3 K −4
59.
60.
Thermal
conductivity (κ )
Stefan’s constant
(σ )
2.9
( Temperature )
4
61.
Wien constant ( b )
Wavelength × Temperature
Km
M 0 LT 0 K
62.
Boltzmann constant
( kB )
Energy
Temperature
JK −1
ML2T −2 K −1
(Continued)
02_Measurements, General Physics_Part 1.indd 9
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2.10 JEE Advanced Physics: Mechanics – I
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
63.
Universal gas
constant ( R )
Pressure × Volume
Mole × Temperature
JK −1mol −1
ML2T −2 K −1mol −1
64.
Charge ( Q )
Current × Time
C (coulomb)
M 0 L0TA
65.
Current density ( J )
Current
Area
Am −2
M 0 L−2T 0 A
66.
Voltage, electric
potential,
electromotive force
( V or E )
Work
Charge
V (volt)
ML2T −3 A −1
67.
Resistance ( R )
Potential difference
Current
ohm ( Ω )
ML2T −3 A −2
68.
Capacitance ( C )
Charge
Potential difference
farad (F)
M −1 L−2T 4 A 2
69.
Electrical
resistivity ( ρ ) or
Resistance × Area
Length
Ωm
ML3T −3 A −2
NC −1
or Vm −1
MLT −3 A −1
⎛ Electrical ⎞
⎜⎝ conductivity ⎟⎠
−1
70.
Electric field ( E )
Electrical force
Charge
71.
Electric flux ( ϕE )
Electric field × Area
NC −1m 2
(or Vm)
ML3T −3 A −1
72.
Electric dipole
moment ( p )
Torque
Electric field
Cm
(Coulomb
metre)
M 0 LTA
73.
Electric field
strength or electric
intensity ( E )
Potential difference
Distance
Vm −1
MLT −3 A −1
74.
Magnetic field,
magnetic flux
density, magnetic
induction ( B )
Force
Current × Length
tesla (T)
ML0T −2 A −1
Magnetic flux ( ϕB )
Magnetic field × Area
75.
(
weber
( = Tm 2 )
)
ML2T −2 A −1
(Continued)
02_Measurements, General Physics_Part 1.indd 10
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2.11
Chapter 2: Measurements and General Physics
Sl.
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
SI Unit
Dimensional
Formula
76.
Inductance ( L )
Magnetic flux
Current
henry (H)
ML2T −2 A −2
77.
Magnetic dipole
moment ( m
)
Torque
or Current × Area
Magnetic field
Am 2
M 0 L2T 0 A
78.
Magnetic field
strength, magnetic
intensity or
magnetic moment
density ( B or H )
Magnetic moment
Volume
Am −1
M 0 L−1T 0 A
Permittivity constant
(of free space) ( ε 0 )
Charge × Charge
C 2 N −1m −2
M −1 L−3T 4 A 2
NA −2
MLT −2 A −2
79.
4π × Electric force × ( Distance )
2
80.
Permeability
constant
(of free space) ( μ0 )
2π × Force × Distance
Current × Current length
81.
Refractive index
(μ)
Speed of light in vacuum
Speed of light in medium
82.
Faraday constant
(F)
Avogadro constant × Elementary charge
Cmol −1
M 0 L0T A mol −1
Wave number ( λ )
2π
Wavelength
m −1
M 0 L−1T 0
83.
M 0 L0T 0
84.
Radiant flux,
Radiant power
Energy emitted
Time
W
ML2T −3
85.
Luminosity of
radiant flux or
radiant intensity
Radiant power of radiant flux of source
Solid angle
Wsr −1
ML2T −3
86.
Luminous power
or luminous flux of
source
Luminous energy emitted
Time
W
ML2T −3
87.
Luminous intensity
or illuminating
power of source
Luminous flux
Solid angle
Wsr −1
ML2T −3
88.
Intensity of
illumination or
luminance
Wm −2sr −1
ML0T −3
Luminous intensity
( Distance )2
(Continued)
02_Measurements, General Physics_Part 1.indd 11
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2.12 JEE Advanced Physics: Mechanics – I
Physical Quantity
and Symbol
Relationship with Other Physical Quantities
89.
Relative luminosity
⎛ Luminous flux of a source ⎞
⎜⎝ of given wavelength ⎟⎠
⎛ Luminous flux of peak sensitivity wavelength ⎞
⎜⎝
⎟⎠
( 555 nm ) source of same power
M 0 L0T 0
90.
Luminous efficiency
Total luminous flux
Total radiant flux
M 0 L0T 0
91.
Illuminance or
illumination
92.
Mass defect ( Δm )
93.
Binding energy of
nucleus ( BE )
94.
Sl.
SI Unit
Luminous flux incident
Area
Dimensional
Formula
Wm −2
ML0T −3
kg
ML0T 0
Mass defect × ( Speed of light in vacuum )
J
ML2T −2
Decay constant ( λ )
0.693
Half life
s −1
M 0 L0T −1
95.
Resonant frequency
( Inductance × Capacitance )− 2
Hz
M 0 L0 A0T −1
96.
Quality factor of
Q-factor of coil ( Q )
Resonant frequency × Inductance
Resistance
97.
Power of lens ( P )
( Focal length )−1
98.
Magnification ( m )
Image distance
Object distance
99.
Capacitive reactance
( XC )
( Angular frequency × Capacitance )−1
ohm ( Ω )
ML2T −3 A −2
100.
Inductive reactance
( XL )
( Angular frequency × Inductance )
ohm ( Ω )
ML2T −3 A −2
( Sum of masses of nucleons ) −
( Mass of the nucleus )
2
1
M 0 L0T 0
dioptre (D)
M 0 L−1T 0
M 0 L0T 0
Quantities Having Same Dimensions
Physical Quantities
Dimensional Formula
Frequency, angular frequency, angular velocity, velocity gradient and decay constant
[ M 0 L0T –1 ]
Work, internal energy, potential energy, kinetic energy, torque, moment of force.
[ M1L2T –2 ]
(Continued)
02_Measurements, General Physics_Part 1.indd 12
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2.13
Chapter 2: Measurements and General Physics
Physical Quantities
Dimensional Formula
Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density.
[ M1L–1T –2 ]
Momentum, impulse.
[ M1L1T –1 ]
Acceleration due to gravity, gravitational field intensity.
[ M 0 L1T –2 ]
Thrust, force, weight, energy gradient.
[ M1L1T –2 ]
Angular momentum and Planck’s constant
[ M1L2T –1 ]
Surface tension, Surface energy (energy per unit area).
[ M1L0T –2 ]
Strain, refractive index, relative density, angle, solid angle, distance gradient, relative
permittivity (dielectric constant), relative permeability etc.
[ M 0 L0T 0 ]
Latent heat and gravitational potential.
[ M 0 L2T –2 ]
Thermal capacity, gas constant, Boltzmann constant and entropy.
⎡⎣ ML2T –2Q –1 ⎤⎦
l
,
g
m
,
k
R
, where l = length, g = acceleration due to gravity, m = mass, k = spring
g
[ M 0 L0T 1 ]
constant, R = Radius of earth.
L
,
R
LC , RC where L = inductance, R = resistance, C = capacitance.
[ M 0 L0T 1 ]
q2
V2
, CV 2 where I = current, t = time, q = charge,
t, VIt , qV , LI 2 ,
C
R
L = inductance, C = capacitance, R = resistance
[ ML2T –2 ]
I 2 Rt ,
Symbols
Following table gives the international symbols for
SI units. In addition to the symbols of fundamental
units, the symbols of derived units have also been
included in this table.
Unit
Symbol
Unit
Symbol
ampere
A
weber
Wb
kelvin
K
ohm
ohm W
candela
cd
volt
V
Unit
Symbol
Unit
Symbol
radian
rad
farad
F
metre
m
joule
J
steradian
sr
henry
H
kilogram
kg
watt
W
newton
N
siemen
S
second
s
coulomb
C
hertz
Hz
tesla
T
(Continued)
02_Measurements, General Physics_Part 1.indd 13
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2.14 JEE Advanced Physics: Mechanics – I
Following points must be noted regarding the
­symbolic representation of various units.
(a) Small letters are used as symbols of units.
­However, if the symbol is derived from a proper
name, then capital letter is used. As an example,
the symbol of newton is “N” and not “n”. It may
be pointed out here only that if a unit is derived
from the name of a person, only the symbol will
be represented by capital letter. The unit itself will
start with small letter. Thus, the unit of force will
be written as “newton” and not as “­Newton”.
(b) The symbols of units are regarded as algebraic
symbols. They are not followed by full stop, dots,
dashes etc. Thus, SI unit of impulse will be represented by Ns and not N-s or N s.
(c) Some space is always left between the number
and the symbol of the unit. Thus, it will be incorrect to write 2.4 kg. The correct representation is
2.4 kg.
(d) Even if the unit is in the plural form, “s” is not
mentioned at the end of the unit. The same is true
for its symbolic representation. Thus, it will be
incorrect to write “metres”. This will be written
as “metre”. In the same way, the symbol will be
m and not ms.
Principle of Homogeneity and Uses
of Dimensional Analysis
According to this principle, the dimensions of all the
terms of the two sides of an equation must be the
same. If
2
X = A ± ( BC ) ±
DE
F
then according to Principle of Homogeneity
Conceptual Note(s)
(a) Suppose in any formula, ( L ± α ) term is coming (where L is length). As length can be added
only with a length, so α should also be a kind of
length.
⇒ [ α ] = M0L1T 0 = L
(b) Similarly consider a term ( F ± β ) where F is
force. A force can be added/subtracted with a
force only and give rise to force. So β should be a
kind of force and its result ( F ± β ) should also be
a kind of force.
F±β
So, only like Physical Quantities (scalars with scalars and vectors with vectors) can be added or
subtracted. However there is no restriction on
multiplication and division of Physical Quantities.
Illustration 1
Check
the
dimensional
1
1
Fs = mv 2 − mv02 .
2
2
1
1
mv 2 − mv02 2
2
According to Principle of Homogeneity
Given
⎡ DE ⎤
a b c
⎥=M LT
⎣ F ⎦
(a) check the dimensional correctness of a physical
relation.
(b) convert units from one system to another.
(c) find dependency of a physical quantity on other
physical quantities.
02_Measurements, General Physics_Part 1.indd 14
correctness
of
Solution
Fs =
1
…(1)
1
[ Fs ] = ⎢⎡ mv 2 ⎤⎥ = ⎡⎢ mv02 ⎤⎥
⎣2
⎦ ⎣2
⎦
[ X ] = [ A ] = ⎡⎣ ( BC )2 ⎤⎦ = ⎢
Taking help from the Principle of Homogeneity
and knowing the dimensional formulae of various
physical quantities, dimensional analysis can be
employed to
β should be a
kind of force.
[ β ] = MLT –2
A force and hence
its dimension will
also be MLT –2
⇒
( MLT −2 ) L = M ( L2T −2 ) = M ( L2T −2 ) ⇒
ML2 T −2 = ML2 T 2 = ML2 T −2 Illustration 2
A student finds that pressure can be expressed as
4 FV 2
P=
, where F is force, V is velocity, t is time
π t2x
and x is distance. However his teacher is doubtful
11/28/2019 6:49:53 PM
Chapter 2: Measurements and General Physics
about the expression. Express the way his teacher
confirms the correctness of expression.
solution
Dimension of LHS = [ P ] = M1 L−1T −2
⎡ 4 FV 2 ⎤ [4][F][v 2 ]
Dimension of RHS is ⎢ 2 ⎥ =
⎣ π t x ⎦ [π ][t 2 ][x]
⎡ 4 FV 2 ⎤ ( M1 L1T −2 )(L2 T −2 )
= M1 L2 T −6
⎢ 2 ⎥=
2
(T )(L)
⎣ πt x ⎦
⇒
⇒
[a]
= M1 L−1T −2
[V 2 ]
[a]
( L3 )2
−1 −1
=M L T
−2
⇒ [ a ] = M1 L5 T −2 and [b] = L3 illustration 4
α
β
= Fv + 2 , then find dimension formula for α
2
t
x
and β , ω here t is time, F is force, V is velocity, x
is distance.
⇒
Dimension of LHS and RHS are not same. So the relation cannot be correct.
If,
illustration 3
solution
For n moles of gas, Vander Waals equation is
a ⎞
⎛
⎜⎝ P + 2 ⎟⎠ ( V − b ) = nRT . Find the dimensions of a
V
⎡ β ⎤
So, ⎢ 2 ⎥ should also be M1 L2 T −3
⎣x ⎦
Since [ Fv ] = M1 L2 T −3
and b , where P is gas pressure. V is volume of gas
and T is temperature of gas.
⇒
solution
⇒
a ⎞
⎛
− b) = nRT
⎜⎝ P + 2 ⎟⎠ × (V
V volume
[ β ] = M1L2T −3
[ x2 ]
[ β ] = M1L4 T −3 β ⎤
⎡
and ⎢ Fv + 2 ⎥ will also have dimension M1 L2 T −3
x ⎦
⎣
pressure
⎡ a ⎤
⇒ [P] = ⎢ 2 ⎥ and [b] = [V ] = L3
⎣V ⎦
2.15
⇒
[α ]
[ t2 ]
= M1 L2 T −3
⇒ [α ] = M 1L2T −1 Test Your Concepts-I
based on Principle of homogeneity and Verification
(Solutions on page H.5)
1. If a composite physical quantity in terms of
moment of inertia I, force F, velocity v, work W
⎛ IFv 2 ⎞
and length L is defined as, Q = ⎜
⎟ . Find the
⎝ WL3 ⎠
dimensions of Q.
2. Can two physical quantities have same dimensions? Explain with example.
3. Find dimensions of universal gas constant R,
universal gravitational constant G.
4. The rate of flow (V) of a liquid flowing through a
⎛P⎞
pipe of radius r and a pressure gradient ⎜ ⎟ is
⎝ ⎠
given by Poiseuille’s equation:
02_Measurements, General Physics_Part 1.indd 15
π Pr 4
V=
8 ηl
Check the dimensional consistency of this
equation.
5. Check the correctness of the equation:
y = a sin( ω t + ϕ ) , where y = displacement,
a = amplitude, ω=angular frequency and ϕ is an
angle.
6. If E, M, J and G respectively denote energy, mass,
angular momentum and gravitational constant,
EJ2
calculate the dimensions of 5 2 .
MG
11/28/2019 6:50:00 PM
2.16 JEE Advanced Physics: Mechanics – I
7. In the formula x = 3 yz 2, x and z are the dimensions
of capacitance and magnetic induction respectively. Find the dimensions of y in MKSQ system.
8. State whether the following statement is true or
false. Give very brief reason in support of your
answer.
e2
The quantity
is dimensionless. Here e, h
2ε 0 hc
and c are electronic charge, Planck’s constant and
velocity of light respectively and ε 0 is the permittivity constant of free space.
9. When white light travels through glass, the refractive index of glass (m = velocity of light in air/­velocity
of light in glass) is found to vary with wavelength as
Conversion of Units from One
System to Another
The measure of a physical quantity is given by
If a physical quantity X has dimensional formula
M a Lb T c and if (derived) units of that physical quantity in two systems are M1a Lb1T1c and M2a Lb2 T2c , respec-
tively and n1 and n2 be the numerical values in the
two systems respectively, then
n1 [ u1 ] = n2 [ u2 ]
(
)
(
)
n1 M1a Lb1T1c = n2 M2a Lb2 T2c a
⇒
Illustration 5
If 1 hp is 746 watt, comment on the statement “1 hp
is 550 ft lb/s”.
Solution
nu = constant
⇒
B
. Using the particle of homogeneity of
λ2
dimensions, find the SI unit in which the constants
A and B must be expressed.
10. A man walking briskly in rain with speed v must
slant his umbrella forward making an angle q with
the vertical. A student derives the following relation between q and v as tanθ = v and checks that
the relation has a correct limit: as v → 0 , θ → 0 ,
as expected. (We are assuming there is no strong
wind and that the rain falls vertically for a stationary man). Do you think this relation can be correct?
If not, guess at the correct relation.
μ = A+
b
c
⎛M ⎞ ⎛L ⎞ ⎛T ⎞
n2 = n1 ⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ 1 ⎟
⎝ M2 ⎠ ⎝ L2 ⎠ ⎝ T2 ⎠ a , b and c are the respective dimensions in mass,
length and time of the physical quantity to be
converted.
M1 , L1 and T1 are fundamental units of mass, length
and time in the first (known) system and M2 , L2 and
T2 are fundamental units of mass, length and time in
the second (unknown) system.
Thus knowing the values of fundamental units in
two systems and numerical value in one system, the
numerical value in other system may be evaluated.
02_Measurements, General Physics_Part 1.indd 16
hp is unit of power, so dimensional formula will be
ML2 T −3 . Hence a = 1 , b = 2 and c = −3 .
So 1 hp = 746 watt = 746 kg m 2 s −3
Now M1 = 1 kg M2 = 1 lb
L1 = 1 m L2 = 1 ft
T1 = 1 sec T2 = 1 sec
n1 = 746 n2 = ?
a
b
⎛ M1 ⎞ ⎛ L1 ⎞ ⎛ T1 ⎞
n2 = n1 ⎜⎝ M ⎟⎠ ⎜⎝ L ⎟⎠ ⎜⎝ T ⎟⎠
2
2
2
c
where a = 1 , b = 2 , c = −3
1
2
⎛ 1 kg ⎞ ⎛ 1 m ⎞ ⎛ 1 sec ⎞
n2 = 746 ⎜
⎟
⎜
⎟ ⎜
⎝ 0.4536 kg ⎟⎠ ⎝ 0.3048 m ⎠ ⎝ 1 sec ⎠
⇒
n2 = 746 × 23.73 {as 1 lb = 0.4536 kg, 1 ft = 0.3048 m}
⇒
−3
n2 = 17702.55 As n1u1 = n2 u2
1 hp = 746 watt = 17702.55 lbft 2 s −3
11/28/2019 6:50:07 PM
Chapter 2: Measurements and General Physics
⎛ ft ⎞
Since (1 lb) × ⎜ 1 2 ⎟ = 1 poundal
⎝ s ⎠
17702.55 lbft 2 s −3 = 17702.55
foot-poundal
sec
Since 1 lb = 32.2 poundal
M=
⇒
M = 13600 × ( 9.183 ) × 10 45 ⇒
M = 1.29 × 10 51 kg ( 9.183 × 1015 )−3 3
L = 9.183 × 1015 m,
⇒ 1 hp ≈ 550 ftlbs −1 T = 3.061 × 107 s
Illustration 7
Conceptual Note(s)
The poundal is a non-SI unit of force and is a part of
(lb)(ft)
.
FPS system of units. It is equal to 1
s2
In two systems of relations between velocity, acceleration and force are respectively given by
v2 =
F
α2
v1 , a2 = αβ a1 and F2 = 1 .
β
αβ
If α and β are constants then find the relations
between mass, length and time in two systems.
Illustration 6
(
)
If velocity of light in air = 3 × 108 ms −1 , acceleration
(
)
Solution
due to gravity = 9.8 ms −2 and density of mercury at
0°C 13600 kgm −3
v2 = v1
) be chosen as fundamental units,
find the unit of mass, length and time.
⇒
Solution
2
( L2T2−1 ) = ( L1T1−1 ) αβ ( L2T2−2 ) = ( L1T1−2 ) αβ LT −1 = 3 × 108 ms −1 …(1)
⇒
LT −2 = 9.8 ms −2 …(2)
ML−3 = 13600 kgm −3 Since F2 =
…(3)
⇒
Dividing (1) by (2)
LT −1
LT
=
−2
α2
β
…(1)
a2 = a1αβ
Given
⇒
13600
⇒
So, M = 1.29 × 10 51 kg,
17702.55 ftlb
⇒ 1 hp =
32.2
s (
2.17
3 × 108
9.8
( M2 L2T2−2 ) = ( M1L1T1−2 ) αβ1 M2 =
T = 3.061 × 10 Substituting for T in equation (1)
−1
= 3 × 108 ⇒
L = 3 × 3.061 × 1015 ⇒
L = 9.918 × 1015 m Substituting for L in equation (3)
M ( 9.183 × 10
F1
αβ
…(3)
Dividing equation (3) by equation (2) we get
7
L ( 3.061 × 107 )
…(2)
)
15 −3
02_Measurements, General Physics_Part 1.indd 17
= 13600 M1
M
= 2 12
(αβ ) αβ α β
Squaring equation (1) and dividing by equation (2)
we get
L2 = L1
α3
β3
T2 = T1
α
β2
Dividing equation (1) by equation (2) we get
11/28/2019 6:50:14 PM
2.18 JEE Advanced Physics: Mechanics – I
Test Your Concepts-II
based on Principle of homogeneity: conversion
(Solutions on page H.5)
1. Calorie is the unit of heat energy and its value is
4.18 J. Suppose we use a new system of units in
which the unit of mass is α kg, unit of length is β m
and that of time is γ s. Find the value of calorie in
terms of the new system of units.
2. Express the power of 100 W bulb in CGS unit with
proper prefix.
3. The CGS unit of viscosity is poise (P). Find how
many poise are there in 1 MKS unit of viscosity
called poiseuille (PI)?
4. If the units of force, energy and velocity are 20 N,
200 J and 5 ms-1, find the units of length, mass
and time.
5. Density of a material in the cgs system is 8 gcc-1.
In a system of units in which the unit of mass is
20 g and that of length is 5 cm, what is the density
of the material in this new system of units.
to DeriVe the neW relations
If a physical quantity X depends on other physical
quantities P , Q and R (say), then we may write
a b c
…(1)
X ∝P Q R
respectively.
Then writing dimensional formula for X , P , Q and
R and equating the dimensions on either sides give
the values of a, b and c . The substitution of these
values in (1) gives the new dimensional relation.
illustration 8
6. Given that 1 pound = 1 lb = 0.4536 kgwt, 1 foot =
0.3048 m, then by dimensional analysis find
the value of 1 horse power. Given that 1 hp =
550 foot lbs-1.
7. Calculate the dimensions of linear momentum and
surface tension in terms of velocity v, density ρ and
frequency ν as fundamental quantities.
8. A new unit of length is chosen such that the speed
of light in vacuum is unity. What is the distance
between the sun and earth in terms of the new unit
if light take 8 min and 20 s to cover this distance.
9. If the unit of force is 1 kN, unit of length 1 km and
the unit of time is 100 s, what will be the unit of
mass.
10. It is estimated that per minute each cm2 of earth
receives about 2 calorie of heat energy from the
sun. This constant is called solar constant S. Express
solar constant in SI units.
If the function is product of power functions of l, m,
g, θ
a
b c d
T = kl m g θ
where k is dimensionless constant
T = ( L ) ( M ) ( LT −2 )
a
b
[ T ] = [ M b La + c T −2c ]
⇒
T = M b La + c T −2c { θ is dimensionless}
Equating the exponents of similar quantities, we get
b = 0 ; a + c = 0 ; −2c = 1
The time period ( T ) of a simple pendulum depends
upon the length of the thread ( l ) , mass of bob ( m ) ,
acceleration due to gravity ( g ) and the angle of
swing ( θ ) . Find the relation of t with other physical
quantities.
⇒
solution
on experimental grounds
It is found experimentally that T depends upon
length of thread ( l ) , mass of bob ( m ) , acceleration
due to gravity g , and angle of swing ( θ ) .
So T = f ( l , m, g , θ )
K = 2π
⇒
T = 2π
02_Measurements, General Physics_Part 1.indd 18
c
Solving for a , b , c we get
a=
1
1
, b = 0, c = −
2
2
T = Kl1 2 g −1 2 l
g
11/28/2019 6:50:20 PM
Chapter 2: Measurements and General Physics
From, this illustration, it is clearly that dimensional
analysis does not provide any information about
dependence of a physical quantity on dimensionless
physical quantities (like θ angle of swing).
Illustration 9
Given that the time period t of oscillation of a gas
bubble from an explosion under water depends upon
the static pressure p , the density of water d and the
total energy of explosion E . Find a dimensional relation for t .
⇒
Substituting in (3), to get
⎛ 5⎞
−2 ⎜ − ⎟ − 2c = 1
⎝ 6⎠
⇒
5
− 2c = 1
3
⇒
2c =
5
−1
3
⇒
2c =
2
3
t ∝ p a d b Ec
1
3
Hence, we get
t = kp a db EC a=−
⇒
Solution
where k is a dimensionless constant
⇒
T = ( ML−1T −2 ) ( ML−3 ) ( ML2 T −2 ) ⇒
T = M a + b + c L− a − 3 b + 2c T −2 a − 2c a
b
c
⇒
a+b+c = 0
…(1)
− a − 3b + 2c = 0 …(2)
−2 a − 2c = 1 …(3)
From (3), we get
1
a+c = −
2
Adding (3) and (4), we get
5
−3 a =
2
⇒
a=−
5
6
02_Measurements, General Physics_Part 1.indd 19
5 1 1
6 d2 E3
The planets move round the sun in nearly circular
orbits. Assuming that the period of rotation depends
upon the radius of the orbit, the mass of the Sun and
the Universal Gravitational constant. Show that the
planet obeys the Kepler’s Third Law of Planetary
Motion.
2
3
T ∝ r
Now, according to the problem, we have
1
2
Substituting in (2), we get
3
− a − + 2c = 0
2
3
2
T = kp
−
To show that the planet obeys Kepler’s Third Law of
planetary motion, we have to prove that
b=
− a + 2c =
5
1
1
, b= , c=
6
2
3
Solution
Substituting in (1),
1
− +b = 0
2
⇒
c=
Illustration 10
Using Principle of Homogeneity we get
⇒
2.19
T ∝ r a MSbG c
⇒
T = kr a MSbG c where k is a dimensionless constant.
…(4)
⇒
T = La M b ( M −1 L3 T −2 ) ⇒
T = M b − c La + 3 c T −2c c
Using the Principle of Homogeneity, we get
b−c = 0
…(1)
a + 3c = 0 −2c = 1 …(2)
…(3)
11/28/2019 6:50:27 PM
2.20 JEE Advanced Physics: Mechanics – I
⇒
1
1
3
c=− , b=− , a=
2
2
2
So, we get
…(2)
−2 a − 2c = 0 …(3)
So, 2 × ( 1 ) + ( 3 ) gives
1
3
1
−
−
T = kr 2 MS 2 G 2
⇒
−1 − 3b + c = 1 2b − 2c = 0 …(4)
Again 2 × ( 2 ) + ( 4 ) gives
⎛ k2 ⎞ 3
T2 = ⎜
r
⎝ MSG ⎟⎠
b = −1
Since the product MSG is constant, so
2
3
T ∝ r
⇒
a = 1
⇒
c = −1 {∵ of ( 1 ) }
So, a = 1 , b = −1 and c = −1
Illustration 11
The height to which the liquid rises in the capillary
tube of radius r depends upon in addition to r on
the surface tension S of the liquid, the density d
of the liquid and the acceleration due to gravity g .
Is it possible to obtain dimensionally a relation for
h without the experimental information that h is
inversely proportional to r ? What is the relation?
Solution
No, it would not be possible to obtain dimensionally
a relation for h without the additional experimental
1
information that h ∝ . This is because without the
r
1
information h ∝ , we will get four variables to be
r
calculated from three equations which is just impossible. So, we shall be writing
h ∝ r −1S a db g c
⇒
h = kr −1S a db g c
⇒
L = L−1 ( MT −2 ) ( ML−3 ) ( LT −2 ) ⇒
L = M a + b L−1− 3 b + c T −2 a − 2c b
c
02_Measurements, General Physics_Part 1.indd 20
⇒
⎛ S ⎞
h = k⎜
⎝ rgd ⎟⎠
Illustration 12
If the velocity of light (c), gravitational constant (G)
and Planck’s constant (h) are chosen as fundamental
units, then find the dimensional formula of mass in
this new system.
Solution
Let m ∝ c x G y h z or m = K c x G y h z
By substituting the dimension of each quantity in
both sides
1 0 0
−1 x
−1 3 −2 y
2 −1 z
[ M L T ] = K[LT ] [ M L T ] [ ML T ]
− y + z x+ 3 y + 2z −x −2y − z
1 0 0
L
T
]
[ M L T ] = [ M
By equating the power of M , L and T in both sides:
x=
1
1
1
, y = − and z =
2
2
2
So, m ∝ c1/2G −1/2 h1/2
Using Principle of Homogeneity, we get
a+b = 0
h = kr −1Sd −1 g −1 − y + z = 1 , x + 3 y + 2 z = 0 , − x − 2 y − z = 0
By solving above three equations, we get
where k is a dimensionless constant
a
⇒
…(1)
11/28/2019 6:50:35 PM
Chapter 2: Measurements and General Physics
2.21
Test Your Concepts-III
based on Principle of homogeneity: Dependence
(Solutions on page H.6)
1. Given that the time period of oscillation of a small
drop of liquid under the influence of surface tension depends upon the density d, radius r and the
surface tension S. Find the expression for the time
period.
2. Given that the time period t of oscillation of a gas
bubble from an explosion under water depends
upon the static pressure p, the density of water d
and the total energy of explosion E. Find a dimensional relation for t.
3. The time period of vibration of a stretched string
depends upon the mass m of the string, tension f
in the string and the length of the string. Find an
expression for the time period of oscillation of the
string.
4. A small steel ball of radius r is allowed to fall under
the gravity through a column of liquid of coefficient of viscosity η. After some time the ball attains
a constant velocity called the terminal velocity vT.
This terminal velocity depends upon the weight W,
the coefficient of viscosity η and the radius of the
ball r. Find an expression for the terminal velocity.
5. A liquid is flowing steadily through a capillary
tube. The rate of flow of the volume of the liquid
depends upon the coefficient of viscosity η, radius
of the tube r and the pressure gradient along the
tube p. Find an expression for the rate of flow of
the volume of the liquid through the tube.
liMitations of DiMensional analysis
The following are the limitations of dimensional
analysis
(a) One has to make a guess about the dependence
of the Physical Quantity on other Physical Quantities, which may or may not work.
(b) This method fails to find the dependency on functions other than power functions. As an example
1
we cannot find the relation s = ut + at 2 , using
2
the method of dimensional analysis. However
02_Measurements, General Physics_Part 1.indd 21
6. The critical angular velocity ωc of a cylinder inside
another cylinder containing a liquid at which its
turbulence occurs depends on viscosity η, density ρ and the distance d between the walls of the
cylinder. Find the expression for ωc.
7. The height to which the liquid rises in the capillary tube of radius r depends upon in addition to r
on the surface tension S of the liquid, the density d
of the liquid and the acceleration due to gravity g.
Is it possible to obtain dimensionally a relation for
h without the experimental information that h is
inversely proportional to r? What is the relation?
8. The planets move round the sun in nearly circular orbits. Assuming that the period of rotation
depends upon the radius of the orbit, the mass of
the Sun and the Universal Gravitational constant.
Show that the planet obeys the Kepler’s Third Law
of Planetary Motion.
9. If density ρ, acceleration due to gravity g and
frequency ν are the basic quantities, find the
dimensions of force.
10. If velocity, force and time are taken to be fundamental quantities find dimensional formula for
(a) Mass, and
(b) Energy.
dimensional analysis does help us to check the
physical correctness of this relation.
(c) This method would also fail to find the dependency of a Physical Quantity on the two Physical
Quantities with identical dimensional formulae.
(d) It gives no information about dimensionless constants which are to be calculated either by experiments or by actual derivation.
(e) It gives no information about the dependence
of a physical quantity on trigonometrical, exponential and logarithmic functions as all are
dimensionless i.e. it cannot find dependence on
dimensionless physical quantities.
11/28/2019 6:50:35 PM
2.22 JEE Advanced Physics: Mechanics – I
(f) If a physical quantity of mechanics depends on
more than three physical quantities all having
dimensional formulae, then dimensional analysis cannot be used to derive their relationship.
(g) Dimensional correctness does not establish
numerical correctness but reverse is true.
Conceptual Note(s)
Consider the terms, sinq, cosq, tanq (and their
reciprocals) where q is dimensionless and
⎛ Perpendicular ⎞
is also dimensionless.
sinθ = ⎜
⎝ hypotenuse ⎟⎠
Similarly cosθ and tanθ are also dimensionless.
Whatever comes in sin( ..... ) is dimensionless and
entire [ sin( ..... ) ] is also dimensionless. So,
(a)
sin (.....)
dimensionless
(b)
So, to conclude we have
1. sin x , cos x , tan x , cosec x , sec x , cot x , log x , e x , a x
all are dimensionless.
i.e., [ sin x ] = [ cos x ] = [ tan x ] = [ cosec x ] =
[ sec x ] = [ log x ] = [ e x ] = [ a x ] = M0L0 T 0
2. The argument of all the functions i.e. x is also
dimensionless. Hence
[ x ] = M0L0 T 0
Illustration 13
If V
F
is velocity,
Solution
α=
⇒
dimensionless
tan(.....)
dimensionless
(d)
F
V2
sin( β t)
dimensionless
cos(.....)
(c)
is time and
sin ( βt ) , then find the dimensional formula
V2
of α and β .
α=
dimensionless
dimensionless
is force, t
F
⇒
[ βt ] = M 0 L0 T 0 [ β ] = T −1 Similarly [ α ] =
dimensionless
2(........)
⇒
dimensionless
[F]
[V 2 ]
[ M1L1T −2 ]
α=
= M1 L−1T 0
2
1
−
1
[L T ]
Illustration 14
dimensionless
(e)
dimensionless
e(........)
dimensionless
(f)
dimensionless
loge(.....)
dimensionless
02_Measurements, General Physics_Part 1.indd 22
FV 2
⎛ 2πβ ⎞
log e ⎜ 2 ⎟ where F is force, V is
⎝ V ⎠
β
­velocity, then find the dimensional ­formula of α and
β.
If, α =
2
Solution
α=
dimensionless
Fv 2
β2
dimensionless
loge
2 πβ
V2
dimensionless
11/28/2019 6:50:41 PM
Chapter 2: Measurements and General Physics
⇒ [α ] =
[ F ][ V 2 ]
⎡⎣ β 2 ⎤⎦
[ 2π ][ β ]
Also,
(c) Least count (LC) of screw gauge (SG) is given by
…(1)
⇒
⇒
[ β ] = L2T −2 LC =
=1
[V 2 ]
[ 1 ][ β ]
=1
L2 T −2
2.23
Pitch( p )
Number of parts on circular scale ( n )
These have been discussed in detail at the end of
the chapter.
(d)
Measurement from Device
Least Count
Substituting β in (1), we get
2.890 m
0.001 m
( M1L1T −2 ) ( L2T −2 )
( L2T −2 )2
0.005 kg
0.001 kg
15.01 cm
0.01 cm
327.92 mm
0.01 mm
1111.111 m
0.001 m
[α ] =
⇒ [ α ] = M1 L−1T 0 LEAST COUNT
The minimum measurement that can be actually
taken by an instrument is called the least count.
The least count of a metre scale graduated in millimetre mark is 1 mm.
The least count of a watch having second’s hand is
1 second.
1
th of a second
The least count of a stop watch is
100
1
i.e.,
s.
100
Conceptual Note(s)
(a) Least count (LC) of vernier calliper (VC) is given by
⎧ Value of one ⎫ ⎧ Value of one ⎫
⎪ ⎪
⎪
⎪
LC = ⎨ part on main ⎬ − ⎨ part on vernier ⎬
⎪
⎪ ⎪
⎪
scale
scale
⎩
⎭ ⎩
⎭
⇒ LC = 1 MSD − 1 VSD
where
MSD = Main Scale Division
VSD = Vernier Scale Division
(b) Least count (LC) of vernier calliper (VC) is also
given by
LC =
Value of 1 part on main scale ( s )
Number of parts on vernier scale ( n )
02_Measurements, General Physics_Part 1.indd 23
SIGNIFICANT FIGURES
Significant figures in the measured value of a physical
quantity are the number of digits which are known reliably
plus the one additional digit which is uncertain.
Larger the number of significant figures obtained in a
measurement, greater is the accuracy of the measurement. The reverse is also true.
The following rules are observed in counting the
number of significant figures in a given measured
quantity.
RULE-1
All non-zero digits are significant.
EXAMPLE:
42.3 has three significant figures.
243.4 has four significant figures.
24.123 has five significant figures.
RULE-2
All zeros between two non-zero digits are significant.
EXAMPLE:
5.03 has three significant figures.
5.004 has four significant figures.
140.004 has six significant figures.
11/28/2019 6:50:44 PM
2.24 JEE Advanced Physics: Mechanics – I
RULE-3
Leading zeros or the zeros placed to the left of the
number are never significant
EXAMPLE:
0.543 has three significant figures.
0.045 has two significant figures.
0.006 has one significant figures.
RULE-4
Zeros that occur at the end of a number (i.e., on the
right-hand side) without an expressed decimal point
are ambiguous (i.e., we have no information on
whether they are significant or not) and hence they
are not considered to be significant.
EXAMPLE:
575000 has three significant figures.
3000 has one significant figure.
Conceptual Note(s)
(a) The measurements like 190 km may have 2 or 3
significant figures and 50800 calorie may have 3,
4 or 5 significant figures.
(b) The potential ambiguity here can be avoided by
making use of Scientific Notation. Depending on
whether 3, 4 or 5 significant figures are correct,
we can write 50800 calorie as:
5. 08 × 104 calorie has 3 significant figures.
5. 080 × 104 calorie has 4 significant figures.
5. 0800 × 104 calorie has 5 significant figures.
RULE-5
Trailing zeros or the zeros placed to the right of the
number are significant.
EXAMPLE:
4.330 has four significant figures.
433.00 has five significant figures.
343.000 has six significant figures.
RULE-6
In exponential notation (or the results expressed in
powers of 10), the numerical portion gives the number of significant figures i.e., the powers of 10 are not
to be counted as significant figures.
02_Measurements, General Physics_Part 1.indd 24
EXAMPLE:
1.32 × 10 −2 has three significant figures.
1.32 × 104 has three significant figures.
RULE-7
The number of significant figures does not depend
upon the system of units used to measure the quantity.
EXAMPLE:
164 mm, 16.4 cm, 0.164 m, 0.000164 km, 164 × 10 −6 km
all have three significant figures.
RULE-8
There are certain measurements that are exact, like “If
the number of students sitting in a class is say 125”.
Then the number of students sitting in the class is
125.00000000000000000 … and so this type of measurement is infinitely accurate and possesses infinite
significant figures.
EXAMPLE:
Number of apples in a pack is 12,
Number of spheres in a box is 25
Both have infinite significant figures.
ROUNDING OFF
While rounding off measurements, we use the
­following rules by convention:
(a) If the digit to be dropped is less than 5, then the
­preceding digit is left unchanged.
EXAMPLE:
7.82 rounded off to one decimal place (RO1DP) is
7.8
3.94 rounded off to one decimal place (RO1DP) is
3.9
47.833 rounded off to one decimal place (RO1DP)
is 47.8
47.862 rounded off to two decimal place (RO2DP)
is 47.86
(b) If the digit to be dropped is more than 5, then the
­preceding digit is raised by one.
EXAMPLE:
6.87 rounded off to one decimal place (RO1DP) is
6.9
12.78 rounded off to one decimal place (RO1DP) is
12.8
11/28/2019 6:50:44 PM
2.25
Chapter 2: Measurements and General Physics
47.862 rounded off to one decimal place (RO1DP)
is 47.9
6.758 rounded off to two decimal place (RO2DP) is
6.76
(c) If the digit to be dropped is 5 followed by digits other
than zero, then the preceding digit is raised by one.
EXAMPLE:
16.351 rounded off to one decimal place (RO1DP)
is 16.4
6.758 rounded off to one decimal place (RO1DP) is
6.8
(d) If digit to be dropped is 5 or 5 followed by zeros, then
preceding digit is left unchanged, if it is even.
EXAMPLE:
3.250 rounded off to one decimal place (RO1DP) is
3.2
12.6850 rounded off to two decimal place (RO2DP)
is 12.68
12.6850 rounded off to one decimal place (RO1DP)
is 12.7
(e) If digit to be dropped is 5 or 5 followed by zeros, then
the preceding digit is raised by one, if it is odd.
EXAMPLE:
3.750 rounded off to one decimal place (RO1DP) is
3.8
16.150 rounded off to one decimal place (RO1DP)
is 16.2
PRECISION and ACCURACY OF A
MEASUREMENT
Precision
Precision of measurement is the uncertainity in number. The precision of measurement is determined
by the least count of measuring instrument. The
smaller is the least count, larger is the precision of
measurement.
Length = ( 3.3760 ± 0.0005 ) × 107 m
The length 3.376 × 107 has an accuracy of four significant figures and precision of 0.0005 × 107 m.
SIGNIFICANT FIGURES IN CALCULATIONS:
FEW EXAMPLES
In most of the experiments, the observations of various measurements are to be combined mathematically, i.e., added, subtracted, multiplied or divided as
to achieve the final result. Since, all the observations
in measurements do not have the same precision, it
is natural that the final result cannot be more precise
than the least precise measurement. The following
two rules should be followed to obtain the proper
number of significant figures in any calculation.
1. In addition or subtraction of the numbers having different precisions, the result should be
reported to the same number of decimal places
as are present in the number having the least
number of decimal places. In other words, the
final result cannot be more precise than the least
precise measurement. The rule is illustrated by
the following examples:
(i) 33.3
← ( has only one decimal place )
+ 3.11
+ 0.313
answer should be reported⎞
36.723 ← ⎛
⎜⎝ to one decimal place
⎟⎠
(ii)
Accuracy
The accuracy of a measurement is also determined
by the number of significant figures. Larger the
number of significant figures, more accurate is the
measurement.
EXAMPLE:
If you find a length to be between values 3.375 × 107 m
and 3.3765 × 107 m. The standard statement of your
result is:
02_Measurements, General Physics_Part 1.indd 25
(iii)
Answer = 36.7
3.1421
+ 0.241
+ 0.09 ← ( has 2 decimal places )
⎛ answer should be reported ⎞
3.4731 ←
⎜⎝ to 2 decimal places
⎟⎠
Answer = 3.47
62.831 ← ( has 3 decimal places )
− 24.5492
⎛ answer should be reported ⎞
38.2818 ← ⎜ to 3 decimal places after ⎟
⎜
⎟
⎝ rounding off
⎠
Answer = 38.2818
11/28/2019 6:50:47 PM
2.26 JEE Advanced Physics: Mechanics – I
2. In multiplication and division of numbers having different precisions, the final result should
be reported to the same number of significant
figures as that of the original number with minimum number of significant figures. In other
words the result in multiplication or division
cannot be more accurate than the least accurate
measurement.
(i) 142.06
× 0.23 ← ( two significant figures )
32.6738
Answer = 33
(ii) 51.028
×1.31
← ( three significant figures )
66.84668
Answer = 66.8
(iii)
0.90
= 0.2112676
4.26
Answer = 0.21
ORDER OF MAGNITUDE: REVISITED
For determining this power, the value of the quantity
has to be rounded off. While rounding off, we ignore
the last digit which is less than 5. If the last digit is 5
or more than five, the preceding digit is increased by
one. For example,
(a) Speed of light in vacuum = 3 × 108 ms-1 ≈ 108 ms-1
(ignoring 3 < 5 )
(b) Mass of electron = 9.1 × 10-31 kg ≈ 10-30 kg
(as 9.1 > 5 )
Order of Magnitude
In Physics, measurement of all things from atom to
universe is done, so we have to deal with very small
and very large magnitudes. Hence, we often talk
about order of magnitude. In scientific notation the
numbers are expressed as, Number = M × 10 x . Where
M is a number lies between 1 and 10 and x is integer. Order of magnitude of quantity is the power of
10 required to represent the quantity. Order of magnitude is expressed in terms of powers of 10 and is
taken to be 10° if M ≤ 10 and 101 if M > 10
( where 10 = 3.16 ) .
EXAMPLE:
8
−1
Speed of light is 3 × 10 ms , its order of magnitude is
and mass of electron is 9.1× 10 −31 kg, its order of magnitude is 10 × 10 −31 ≈ 10 −30 kg.
Errors in a Repeated Measurement
If we take a measurement experimentally, it necessarily involves errors, due to two factors.
(a) Human errors, which may be due to reaction
time or carelessness.
(b) Experimental errors, which are due to least count
of measuring instruments. For given measuring
instruments, the human errors may be reduced
by repeating experiment for a large number of
times. If a graph is plotted between number of
observations and the observed quantity x, the
graph is shown in fi
­ gure. Such a curve is called
Gaussian distribution or normal distribution
curve.
N
Conceptual Note(s)
(a) Change in units of measurement of a quantity
does not affect the number of significant figures.
(b) Significant figures are quoted for a measurement
not for a pure number.
(c) Greater is number of significant figures in a measurement, smaller is the percentage error.
02_Measurements, General Physics_Part 1.indd 26
x
11/28/2019 6:50:50 PM
2.27
Chapter 2: Measurements and General Physics
MEAN VALUE
Δa1 = aav − a1
If x1 , x2 , x3 ...xn are n measured values of a physical
quantity, then the mean value is given by
Δa2 = aav − a2
………………
………………
N
∑x
x + x2 + x3 + ...xn
x av = x = x = 1
= i =1
N
N
Δan = aav − an
i
The arithmetic mean of all the absolute errors is called
as mean absolute error and is given by
Standard Deviation (σ )
The spread of the experimental data is measured by
the quantity called Standard Deviation, defined as
σ=
1
N −1
∑(x − x )
i
av
2
i
Standard Error in the Mean
The standard error (also called probable error of the
mean) α , for a given set of readings (data) is
α=
σ
N
The larger is the number of readings, the smaller is
the error.
EXAMPLE:
10° × 108 = 108 ms −1
A student takes 100 readings and standard error is e. If
e
e
he takes 400 readings, then the error will be
= . So,
4 2
on taking 400 readings the standard error will be halved.
Absolute Errors
The positive difference between arithmetic mean
value and the measured value in the ith observation
is called as the absolute error of that observation. The
arithmetic mean value is also called as true value.
Absolute error is
( Δa )av = Δa =
02_Measurements, General Physics_Part 1.indd 27
n
1
n
n
∑ Δa
i
Relative and Percentage Error
The relative error is defined as the ratio of the mean
absolute error to the mean value or the true value.
Mathematically,
Relative error =
⇒
( Δa )av
a
=
Δa
a
Percentage relative error =
Δa
× 100%
a
Illustration 15
The length of a rod as measured in an experiment is
recorded as 2.50 m, 2.54 m, 2.49 m, 2.58 m, 2.49 m,
2.57 m respectively. Find the mean/true length, absolute error in each case, mean absolute error and the
percentage error.
Solution
Mean length or true length
aav = a =
Δai = aav − ai
where Dai is absolute error in the ith observation.
Then clearly
Δa1 + Δa2 + Δa3 + ... + Δan
i =1
And if we take the single measurement then the result
of measurement will be a ± Δa = aav + ( Δa )av
( mean value or true value ) − ( ith measured value )
⇒
( Δa )av = Δa =
⇒
a1 + a2 + a3 + a4 + a5 + a6
6
a =
2 ⋅ 50 + 2.54 + 2.49 + 2.58 + 2.49 + 2.57
6
a =
15.17
= 2.528
6
a ≅ 2.53 m 11/28/2019 6:50:54 PM
2.28 JEE Advanced Physics: Mechanics – I
Δa1 = aav − a1 = 2.53 − 2.50 = 0.03
Δa2 = aav − a2 = 2.53 − 2.54 = 0.01
Δa3 = aav − a3 = 2.53 − 2.49 = 0.04
Let ΔA and ΔB the absolute errors in A and B .
Then the values of A and B should be recorded as
A ± ΔA and B ± ΔB . If ΔX be the absolute error in
the final result X , then
X ± ΔX = ( A ± ΔA ) + ( B ± ΔB )
Δa4 = aav − a4 = 2.53 − 2.58 = 0.05
Δa5 = aav − a5 = 2.53 − 2.49 = 0.04
⇒
Δa6 = aav − a6 = 2.53 − 2.57 = 0.04
⇒
Δa1 , Δa2 , Δa3 , Δa4 , Δa5 , Δa6 are the absolute
errors in each case.
Mean absolute error i.e., ( Δa )av = Δa is
( Δa )av = Δa =
Δa1 + Δa2 + Δa3 + Δa4 + Δa5 + Δa6
6
0.03 + 0.01 + 0.04 + 0.05 + 0.04 + 0.04
6
⇒
( Δa )av = Δa =
⇒
( Δa )av = Δa =
0.21
= 0.035
6
So, mean length = ( 2.53 ± 0.035 ) m
Δa
⇒ Percentage error =
× 100%
a
0.035
⇒ % age error =
× 100%
2.53
⇒ % age error = 1.38% COMBINATION OR PROPAGATION
OF ERRORS
The formula used in an experiment may involve
addition, subtraction, multiplication or division etc.
of different quantities measured in the experiment.
There may be some error in each of the reading. But
all the errors will not affect the final result to the same
extent. Different errors will affect the final result differently. So the final result will depend upon the way
these errors are combined through mathematical
operations. We shall calculate the maximum possible
error in all the cases.
When the Result Involves the Sum of Two
Observed Quantities
We suppose that the result X is given as the sum of
two observed quantities A and B, i.e.
X = A + B
02_Measurements, General Physics_Part 1.indd 28
X ± ΔX = ( A + B ) ± ( ΔA + ΔB ) ± ΔX = ± ( ΔA + ΔB ) So, maximum possible error in X is
ΔX = ΔA + ΔB
Thus when two quantities are added, the absolute
error in the final result is the sum of the absolute
errors of the quantities.
When the Result Involves the Difference of
Two Observed Values
Let the result X be given as the different of two
observed quantities A and B i.e.
X = A − B
Let ΔA and ΔB be absolute errors in A and B .
If ΔX is the absolute error in the final result, then
X ± ΔX = ( A ± ΔA ) − ( B ± ΔB )
⇒
X ± ΔX = ( A + B ) ± ( ΔA ± ΔB ) ⇒
± ΔX = ± ( ΔA ± ΔB ) But the error in X will be maximum if
ΔX = ΔA + ΔB
Thus when the two quantities are subtracted, the
absolute error (simply we may call it as error) in the
final result is again the sum of absolute errors of the
two quantities.
When the Result Involves the Product of
Two Observed Quantities
We suppose that X = AB , then
X ± ΔX = ( A ± ΔA ) ( B ± ΔB )
⇒
X ± ΔX = AB ± AΔB ± ΔAB ± ΔA ΔB
Dividing both sides of the above equation with X,
we get
X ± ΔX AB ± AΔB ± ΔAB ± ΔAΔB
=
X
AB
11/28/2019 6:51:03 PM
Chapter 2: Measurements and General Physics
⇒ 1±
ΔX
ΔB ΔA ΔAΔB
= 1±
±
±
X
B
A
AB Since ΔA and ΔB are small, so the product ΔAΔB
can be neglected. So we get
⇒
±
ΔX
ΔA ΔB
=±
±
X
A
B To have the maximum relative error, we get
ΔX ⎞ ⎛
ΔA ΔB ⎞
⎛
±
⎟
⎜⎝ 1 ±
⎟ = ⎜ 1±
X ⎠ ⎝
A
B ⎠
⇒
±
⇒
ΔX ΔA ΔB
=
+
X
A
B Alternative Method: The above result can be derived
as follows:
ΔX
ΔA ΔB
=±
±
X
A
B ⇒
⎛ Percentage ⎞ ⎛ Percentage ⎞ ⎛ Percentage ⎞
⎜ Error in ⎟ = ⎜ Error in ⎟ + ⎜ Error in ⎟
⎜⎝ value of X ⎟⎠ ⎜⎝ value of A ⎟⎠ ⎜⎝ value of B ⎟⎠
If we are to find the absolute error ΔX , then we have
X = AB
log X = log A + log B Differentiating on both the sides
⎛ ΔA ΔB ⎞
ΔX = X ⎜
+
⎟ X = AB
⎝ A
B ⎠
dX dA dB
=
+
X
A
B
Now, if, then ΔX = BΔA + AΔB
⇒
ΔX ΔA ΔB
=
+
and
X
A
B
and if X =
⇒
ΔX
ΔA
ΔB
%=
× 100 +
× 100
X
A
B
When the Result Involves the Quotient
of Two Observed Quantities
A
B
−1
X ± ΔX = ( A ± ΔA ) ( B ± ΔB ) ⇒
ΔX ⎞
ΔA ⎞ −1 ⎛
ΔB ⎞
⎛
⎛
X ⎜ 1±
⎟⎠ B ⎜⎝ 1 ±
⎟
⎟⎠ = A ⎜⎝ 1 ±
⎝
X
A
B ⎠
⇒
ΔX ⎞ A ⎛
ΔA ⎞ ⎛
ΔB ⎞
⎛
X ⎜ 1±
⎟ ⎜ 1±
⎟
⎟ = ⎜ 1±
⎝
X ⎠ B⎝
A ⎠⎝
B ⎠
⇒
ΔX ⎞ ⎛
ΔA ΔB ΔAΔB ⎞
⎛
±
±
⎜⎝ 1 ±
⎟ = ⎜ 1±
⎟
X ⎠ ⎝
A
B
AB ⎠
02_Measurements, General Physics_Part 1.indd 29
Suppose X = kAn , where k is a constant
n
Then X ± ΔX = k ( A ± ΔA )
⇒
ΔX ⎞
ΔA ⎞
⎛
n ⎛
X ⎜ 1±
⎟
⎟ = kA ⎜⎝ 1 ±
⎝
X ⎠
A ⎠
n
n
ΔX ⎛
ΔA ⎞
= ⎜ 1±
⎟
⎝
X
A ⎠ ΔA
1 , so from Binomial Theorem, we get
Since
A
⇒
ΔAΔB
AB
CASE-1
⇒ 1±
A ± ΔA
Then X ± ΔX =
B ± ΔB
Neglecting
A
A ⎛ ΔA ΔB ⎞
+
, then ΔX = ⎜
⎟
B
B⎝ A
B ⎠
In Case of Power Functions
Thus, the final relative error, when the quantity is the
product of two observed quantities is the sum of the
relative errors of the two quantities.
Suppose that X =
So, whether the quantities are being multiplied or
divided, then
ΔX ΔA ΔB
=
+
X
A
B
⇒
⇒
2.29
−1
1±
−1
{∵ ΔA and ΔB are very small }
ΔX
⎛ ΔA ⎞
= ±n ⎜
⎝ A ⎟⎠
X
⇒
±
⇒
ΔX
⎛ ΔA ⎞
= n⎜
⎝ A ⎟⎠
X
ΔX
⎛ ΔA ⎞
= 1± n⎜
⎝ A ⎟⎠
X
{∵ Error in a measurement is always extremum
during measurement}
11/28/2019 6:51:11 PM
2.30 JEE Advanced Physics: Mechanics – I
CASE-2
⎛A B
If X = k ⎜
⎝ C n ⎟⎠
m ⎞
If ΔA , ΔB and ΔC be the respective absolute errors
in calculating the values of A , B and C, then let the
corresponding error in calculating X (depending on
A , B and C ) be ΔX . Then
X ± ΔX =
l
( A ± ΔA ) ( B ± ΔB )
ΔX ⎞
⎛
X ⎜ 1±
⎟=
⎝
X ⎠
ΔA ⎞ m ⎛
ΔB ⎞
⎛
Al ⎜ 1 ±
⎟⎠ B ⎜⎝ 1 ±
⎟
⎝
A
B ⎠
ΔC ⎞
⎛
Cn ⎜ 1 ±
⎟
⎝
C ⎠
ΔC ⎞
⎛
⎜⎝ 1 ±
⎟
C ⎠
l
Conceptual Note(s)
n
ΔA ⎞ ⎛
ΔB ⎞
⎛
⎜⎝ 1 ±
⎟⎠ ⎜⎝ 1 ±
⎟
A
B ⎠
ΔX ⎞ A B
⎛
⇒ X ⎜ 1 ±
⎟=
⎝
X ⎠
Cn
And if X = kA± N , where k is a constant and N is any
real number.
ΔX
ΔA
=N
Then
X
A
m
n
m
⇒
ΔX ⎞ ⎛
ΔA ⎞ ⎛
ΔB ⎞ ⎛
ΔC ⎞
⎛
⎜⎝ 1 ±
⎟⎠ = ⎜⎝ 1 ±
⎟⎠ ⎜⎝ 1 ±
⎟⎠ ⎜⎝ 1 ±
⎟
X
A
B
C ⎠
…(1)
ΔA
ΔB
ΔC
1,
1 and
1 , so from Binomial
A
B
C
Theorem, we have
ΔB
ΔC ⎞
⎛
≅ 1± m
and ⎜ 1 ±
⎟
⎝
B
C ⎠
Illustration 16
If V = ( 50 ± 2 ) V and I = ( 5 ± 0.1 ) A, then find the
percentage error in measuring the resistance. Also
find the resistance with limits of error.
Solution
l
ΔA ⎞
ΔA
⎛
⎜⎝ 1 ±
⎟⎠ ≅ 1 ± l
A
A
m
ΔX
⎛ ΔA ⎞
% = N⎜
× 100%
⎝ A ⎟⎠
X
−n
Since
ΔB ⎞
⎛
⎜⎝ 1 ±
⎟
B ⎠
⎛ %age ⎞
⎛ %age ⎞
⎛ %age ⎞
⎛ %age ⎞
⎜ Error ⎟ = l ⎜ Error ⎟ + m ⎜ Error ⎟ + n ⎜ Error ⎟
⎜⎝ in y ⎟⎠
⎜⎝ in A ⎟⎠
⎜⎝ in B ⎟⎠
⎜⎝ in C ⎟⎠
m
l
ΔX
ΔA
ΔB
ΔC
=l
+m
+n
X
A
B
C
( C ± ΔC )n
l m
⇒
ΔX
ΔA
ΔB
ΔC
= −l
−m
−n
X
A
B
C
From both these relations, we get
−
⇒
m
l
⇒
ΔX
ΔA
ΔB
ΔC
= +l
+m
+n
X
A
B
C
OR
{ l > 0, m > 0, n > 0 and k is a constant }
+
(a) Given V = ( 50 ± 2 ) V and I = ( 5 ± 0.1 ) V
−n
ΔC
≅ 1∓ n
C
We know that R =
V
I
⇒
ΔR ΔV ΔI
2 0.1
=
±
=
+
R
V
I
50 5
ΔB ⎞ ⎛
ΔC ⎞
ΔX ⎛
ΔA ⎞ ⎛
= ⎜ 1± l
⎟⎠ ⎜⎝ 1 ± m
⎟⎠ ⎜⎝ 1 ∓ n
⎟
⎝
X
A
B
C ⎠
⇒
ΔR
2
0.1
( in % ) = × 100 + × 100 = 6 %
R
50
5
ΔX
ΔA
ΔB
ΔC ⎛ Neglected ⎞
= ±l
±m
+
∓n
X
A
B
C ⎜⎝ Terms ⎟⎠ (b) R =
So, (1) becomes
1±
⇒
±
⇒
ΔB
ΔC
ΔX
ΔA
±
= ±l
±m
∓n
X
A
B
C Since, we know that during propagation errors are
always taken to be the extremum, so we have
02_Measurements, General Physics_Part 1.indd 30
V 50
=
= 10 ohm
I
5
ΔR ΔV ΔI
2 0.1
=
+
=
+
V
I
50 5
R
2
0.1
⇒ ΔR = 50 × 10 + 5 × 10 = 0.6
⇒
R ± ΔR = ( 10 ± 0.8 ) ohm
11/28/2019 6:51:19 PM
Chapter 2: Measurements and General Physics
Problem Solving Technique(s)
(j) Errors never propagate in case of constants.
EXAMPLE
(a) Δy is always positive i.e. Δy > 0.
4
If V = π r 3 , then
3
(b) Δy has units same as that of y.
(c) A quantity in terms of absolute error is expressed as
y = ( yt ± Δy ) units
(d) If least count is not given and a measurement is
given, then error in the measurement will be ±1
in last digit.
EXAMPLE
If L = 5.216 metre,
then ΔL = ± 0.001 metre
Also, if M = 2.50 Kg,
1
Δy
⎛ Δx ⎞
= n⎜
⎝ x ⎟⎠
y
irrespective of value of k.
{
Only when
Δx
<< 1
x
}
VERNIER CAlLIPER
Vernier Calliper is a device used to measure the
­internal or external diameter as well as the depth of
a vessel. The maximum precision up to which a scale
can measure is 1 mm. For measuring lengths that
are less than 1 mm, vernier calliper is used. Vernier
­calliper is also called as Slide Calliper.
A Vernier calliper consists of two scales (See Figure)
(a) A Main Scale: It is calibrated in cm on one side
and in inches on the other side. The value of one
main scale division (MSD) is 1 mm or 0.1 cm.
(b) A Vernier Scale: It is the movable scale that can
slide on the main scale. Generally 10 vernier scale
divisions (VSD) is equal to 9 main scale divisions
(MSD). Therefore the value of one vernier scale
9
division (VSD) is equal to
mm .
10
5
4
5
6
7
8
9
E
10 cm
10
3
1
02_Measurements, General Physics_Part 1.indd 31
3
2
0
A
Δr
<< 1
r
D
2
0
}
Only when
n
(k) In general if y = kx , then
⎛ Δy
⎞
y = yt units ± ⎜
× 100 ⎟ %
⎝ yt
⎠
C
{
ΔV 3Δr
=
V
r then ΔM = ± 0.01 kg
(e) The error in a measurement is equal to the least
count of the instrument.
(f) For evaluating error in a formula
(i) the powers are changed into multiplication.
(ii) the multiplication and divisions are changed
into addition.
(g) In case of addition and subtraction, it is advisable
to calculate absolute error first and then relative
error.
(h) In case of multiplication, division and power functions, it is advisable to calculate relative error first
and then absolute error.
(i) Relative Error or Percentage relative error is
dimensionless, hence a measurement can be
expressed in terms of percentage relative error as
2.31
5
B
6
4
1 - Used to take external measures of objects
2 - Used to take internal measures of objects
3 - Used to measure the depth of objects: Depth Probe
4 - Main scale (cm)
5 - Vernier (cm)
6 - Used to block movable part: Retainer
11/28/2019 6:51:21 PM
2.32 JEE Advanced Physics: Mechanics – I
The upper two jaws C and D are used to find the internal diameter of a pipe, test tube and that of hollow
cylinder. The lower two jaws A and B are used to
measure the length and diameter of a cylinder or the
diameter of a sphere by placing the object between
the lower jaws. A thin metallic strip E projects at one
end of the calliper which is used to measure the depth
of the vessel. The smallest length up to which the vernier calliper can measure is called Least Count (LC)
or Vernier Constant.
Least count of vernier calliper is defined as
the ratio of the value of one MSD to the number of
divisions on the vernier scale called Vernier Scale
Division (VSD)
Value of 1 MSD
S
Least count =
=
Number
of divisions on ⎞
N
⎛
⎜⎝
⎟⎠
the vernier scale
Least count of vernier callipers is also defined as the
difference between the value of one MSD and the
value of one VSD.
⇒ Least count = 1 MSD – 1 VSD
It is observed that N VSD (divisions of vernier scale)
coincide with ( N − 1 ) MSD (divisions of the main
scale)
N VSD = ( N − 1 ) MSD
⇒ 1 VSD =
( N − 1)
N
MSD
The zero of the vernier scale lies to the right of zero of
the main scale. When the two jaws are in contact, the
distance between them is zero but the reading is positive. In other words, the instrument measures more
than the actual distance and therefore zero error is positive. Figure A represents the case of positive zero error.
0
1
2
3
4
0 5 10
Figure A
Positive zero error
Negative Zero Error
The zero of vernier scale lies to the left of zero of the
main scale. When the two jaws are in contact, the distance between them is zero but now the reading is
negative. In other words, the instrument measures
less than the actual distance and therefore zero error
is negative. Figure B represents the case of negative
zero error.
0
1
2
3
4
0 5 10
Figure B Negative zero error
Calculating Zero Error
Least count = 1 MSD – 1 VSD
Least Count = 1 MSD −
Positive Zero Error
( N − 1)
MSD =
N
1
MSD
N
To calculate the zero error the two jaws are kept in contact and the number of vernier scale divisions coinciding with some main scale division is noted. For the
arrangement shown in Figure A, the fourth vernier
division coincides with some division of main scale.
Concept of Zero Error
⇒ Zero error = + 4 × (LC) = + 4 × 0.01 = + 0.04 cm When the lower two jaws touch each other, the zero
of the vernier scale coincides with the zero of the
main scale. When this is the case then there is no zero
error. However, if the zero of the vernier scale does
not coincide with zero of the main scale (when the
jaws are in contact), then error exists and necessary
correction has to be applied.
⇒ Zero correction = −0.04 cm 0
1
2
3
⇒ Zero error = − 5 × (LC) = − 5 × 0.01 = − 0.05 cm ⇒ Zero correction = +0.05 cm So, the correct length is
0 5 10
No zero error
02_Measurements, General Physics_Part 1.indd 32
4
For the arrangement shown in Figure B, fifth vernier
division coincides with some main scale division. To
calculate zero error, the number of division to be considered will be 5.
Correct Length = ( Observed Length ) ± ( Zero Error )
11/28/2019 6:51:25 PM
Chapter 2: Measurements and General Physics
Steps to Be Followed While Taking
Readings with Vernier Callipers
(a) Find the Least Count (LC).
(b) For taking the Main Scale Reading (MSR), the
division next before the zero of the vernier
should be considered.
(c) The division of the vernier scale which coincides
with any main scale division should be noted.
This is called Vernier Coincidence (VC).
(d) The Correct Reading (CR) is given by
CR = MSR + ( LC ) × ( VC )
(e) Prefer using a lens to note Vernier Coincidence
(VC).
(f) Vernier principle is also used for measurement of
fraction of angles as in sextant.
this observation, the length of the object under measurement is 2.2 cm plus some fraction (say y ).
STEP-4: The value of the fraction is found by locating the vernier division which coincides with a main
scale division. In present case it is 7. Therefore, fraction to be added is given by
Fraction = 7 × V.C. = 7 × 0.01 cm = 0.07 cm
Hence, the reading of the vernier callipers is
2.2 + 0.07 i.e., 2.27 cm.
STEP-5: So, the correct reading is given by
CR = MSR + ( LC ) × ( VC )
Reading = main scale + vernier scale
= 3.1 cm + 0.04 cm
= 3.14 cm
Precautionary Measures
(a) As a precaution, the jaws (lower) of the callipers
should not be pressed too hard on the object
placed between them so as to get the actual
reading.
(b) Also at any position, the diameter should be
measured in two directions at right angles to
each other.
(c) Same units should be used while calculating the
result.
(d) Vernier coincidence must be noted without any
parallax error by repeating the observations five
times at different positions of the object. Slide
calliper is called vernier callipers since it was
first designed by French mathematician named
Vernier.
How to Measure
STEP-1: Measure the least count/vernier constant.
STEP-2: Place the object between the jaws and tighten
them
0
1
2
3
4
0 5 10
STEP-3: Suppose that during this observation, zero
of the vernier scale (­having vernier constant 0.01 cm)
lies between 2.2 cm and 2.3 cm mark. Therefore, in
02_Measurements, General Physics_Part 1.indd 33
2.33
0
1
2
3
4
0
5 10
The fourth vernier mark coincides with a marking on
the main scale. This gives a reading of 0.4 mm or
0.04 cm to be added to main scale reading.
Illustration 17
Least count of a vernier calliper is 0.01 cm. When the
two jaws of the instrument touch each other the 5th
division of the vernier scale coincide with a main scale
division and the zero of the vernier scale lies to the
left of the zero of the main scale. Further more while
measuring the diameter of a sphere, the zero mark of
the vernier scale lies between 2.4 cm and 2.5 cm and
the 6th vernier division coincides with a main scale
division. Calculate the diameter of the sphere.
Solution
The instrument has a negative error,
e = ( −5 × 0.01 ) cm = −0.05 cm
Measured reading = ( 2.4 + 6 × 0.01 ) = 2.46 cm
True reading = Measured reading −e
⇒ True Reading = 2.46 − ( −0.05 ) ⇒ True Reading = 2.51 cm Therefore, diameter of the sphere is 2.51 cm
11/28/2019 6:51:27 PM
2.34 JEE Advanced Physics: Mechanics – I
SCREW GAUGE
Base Line
A Screw Gauge is a device used to determine thickness (or diameter) of thin sheet or a wire. With its
help one can measure the diameter of very thin wires
or similar objects. Its measures accurately up to
0.001 cm and hence is also commonly called micrometer screw gauge.
A reference line or Base Line or Datum Line is graduated in mm and is drawn on the sleeve cylinder parallel to the axis of nut. It is commonly called Main
Scale or Sleeve Scale or Pitch Scale.
cross-section
of wire
anvil
sleeve
cylinder
5
B
U-frame
The hollow cylinder that moves over the sleeve cylinder is tapered from one end. On the tapered end
graduations are made, which are either 50 or 100 in
number. The scale marked on sleeve is called circular
scale or thimble scale or head scale.
thimble
spindle
0
A
thimble scale
(circular scale)
0
45
40
35
30
main scale
(pitch scale)
datum line
(base line)
Circular Scale or Thimble Scale or
Head Scale
Ratchet
ratchet
Construction
The ratchet is attached to screw with the help of a
spring. When the flattened end B of the screw comes
in contact with the stud A , the ratchet becomes free
and makes a rattling noise. Thus, end B of the screw
will not be further pushed towards the stud A .
The screw gauge consists of the following parts
Pitch of Screw
U-Frame
The pitch of screw is defined as, the distance between
two consecutive threads of screw when measured
along the axis of screw
It is a steel frame having a U shape. On one end of U
shaped frame a screw is fixed permanently. This fixed
screw is commonly called Stud or Anvil and forms
the fixed jaw of the screw gauge. On the other end of
U shaped frame is fixed a Nut through which slides
a screw. The end A of the screw forms the movable
jaw of screw gauge.
Nut and Screw
The nut (at B ) is threaded from inside and the screw
is threaded from outside. The screw can move in and
out of nut by circular motion.
Thimble or Circular Cylinder
The screw is connected to a hollow circular cylinder(s)
also called Thimble, which rotates along with the nut
on turning.
OR
Pitch of screw can also be defined as the forward
distance travelled by the tip of screw (end B ) when
head of screw completes one rotation.
Principle of Screw Gauge
It works on fundamental screw principle that rotational
motion can be converted into translational motion.
Determination of Pitch of Screw
To determine the pitch of the screw gauge, the screw
is given five complete rotations. The distance moved
by the thimble on the main scale is then recorded. The
pitch is calculated by the formula
Sleeve Cylinder
To the nut is attached a hollow cylinder, commonly
called Sleeve Cylinder. The spindle of the screw
passes through this sleeve cylinder.
02_Measurements, General Physics_Part 1.indd 34
⎛ Distance moved by thimble ⎞
⎜
⎟
on Main Scale ( MS )
Pitch = ⎜
⎟
⎜ Number of rotations of thimble ⎟
⎝
⎠
11/28/2019 6:51:29 PM
Chapter 2: Measurements and General Physics
Illustration 18
If 5 mm is the distance moved by the thimble on the
main scale for 5 rotations then calculate the pitch.
If there is no zero error then the reading will be
0.00 mm, as shown. In this case, the zero of the circular scale coincides with Base Line (or Reference Line).
Solution
For a screw gauge, the least count is the smallest distance moved by its tip when the screw turns through
division marked on it.
Determination of Least Count
First of all we shall determine the pitch and also count
the number of divisions on c­ ircular scale. Then least
count is determined by the formula
Least Count =
Pitch
Number
of
divisions on ⎞
⎛
⎜⎝ the Circular Scale (CS) ⎟⎠
Positive Zero Error with Negative
Correction
The zero of circular scale may be left behind the Base
Line or below the Base Line (or Reference Line) as
shown in figure.
In such a case when the studs are in contact (zero
distance), the screw gauge is giving a positive reading. Such a screw gauge measures more and therefore it has got Positive Zero Error or Negative Zero
Correction.
From figure we note that the zero of circular scale has
been left behind by 3 circular divisions.
For a screw gauge having a pitch of 1 mm and having
100 divisions marked on its thimble, the least count is
Least Count =
1 mm
= 0.01 mm = 0.001 cm
100
BACKLASH ERROR
0
Base line
5 mm
Pitch =
= 1 mm
5
Least Count of Screw
2.35
Base line
5
0
Zero below the
base line
So, zero error is
⎛ Zero ⎞
⎛ Number of divisions ⎞ ⎛ Least ⎞
⎟⎠ × ⎜⎝ Count ⎟⎠
⎜⎝ Error ⎟⎠ = + ⎜⎝
left behind
This type of error is associated with all instruments
based on the Screw Principle. In the screw gauge, the
screw moves in a nut. If the screw is not fitting properly in the grooves (threads), then there is irregularity in the movement of the screw, when the head is
rotated. Error due to this is called backlash error. In
order to avoid this error, during an observation the
screw is always moved in one direction.
So, to conclude we have
Determination of Zero Error
Correct Length = Observed Length − Zero Error
On bringing the screw end B in contact with the stud
A , if the zero of the main scale (MS) does not coincide with zero of circular scale (CS), the micrometer
or the screw gauge is said to have zero error.
Negative Zero Error with Positive
Correction
No Zero Error
Before using the screw gauge, we must check for a
zero error. Close the screw gauge so that the spindle
touches the anvil.
02_Measurements, General Physics_Part 1.indd 35
⇒
Zero Error = +3 × 0.001 cm = +0.003 cm ⇒ Zero Correction = −0.003 cm OR
Zero ⎞
⎛ Correct ⎞ ⎛ Observed ⎞ ⎛
⎜⎝ Length ⎟⎠ = ⎜⎝ Length ⎟⎠ + ⎜⎝ Correction ⎟⎠
When the zero line, marked on circular scale lies
above the Base Line of the main scale, then there is a
negative error and the correction is positive. The zero
of circular scale may cross the reference line as shown
in figure. In such a case, if we move the screw back so
as to make the zero of circular scale to coincide with
11/28/2019 6:51:31 PM
2.36 JEE Advanced Physics: Mechanics – I
reference line, a gap will be left between the studs.
Thus such a screw gauge gives the zero reading even
when there is gap between studs i.e., it measures less.
Therefore, the screw gauge possesses negative zero
error or positive zero correction.
Base line
0
45
0
5
0
45
40
35
30
Zero above the
base line
reading = sleeve reading + thimble reading
= 17 × 0.5 mm + 0.40 mm = 8.90 mm
thimble scale has 50
divisions, each of
which is equal to
0.01 mm
To determine the zero correction, note the number
of divisions through which the zero mark of circular
scale has crossed the reference line. From figure, we
note that the number of such divisions that cross the
zero mark is 3.
So, Zero error is
17 divisions
⎛ Zero ⎞
Number of ⎞ ⎛ Least ⎞
⎛
⎜⎝ Error ⎟⎠ = − ⎜⎝ divisions crossed ⎟⎠ × ⎜⎝ Count ⎟⎠
⇒ Zero Error = −3 × 0.001 cm = −0.003 cm Conceptual Note(s)
⇒ Zero Correction = +0.003 cm 0
0
Reading a Screw Gauge
Place the object say, a piece of wire (whose diameter is to be found) between the two studs of the
screw gauge. Turn the screw with the help of ratchet
arrangement, till the ratchet becomes free with the
sound of a click. It ensures the gentle placing of the
wire between the studs.
Let the complete main scale reading (MSR) be read as
a and the nth circular scale division (CSD) coinciding with the reference line.
Then, diameter of the wire,
d = a × pitch + n × least count
Here a = 17 , pitch = 0.5 mm , n = 40 ,
Correct Reading = MSR + ( CSR × LC ) + ZC
OR
Correct Reading = MSR + ( CSR × LC ) − ZE
where, MSR is Main Scale Reading
CSR is Circular Scale Reading, LC is Least Count
ZC is Zero Correction and ZE is Zero Error
Illustration 19
The pitch of a screw gauge is 1 mm and there are 100
divisions on circular scale. When faces A and B are
just touching each without putting anything between
the studs 32nd division of the circular scale coincides
with the reference line. When a glass plate is placed
between the studs, the linear scale reads 4 divisions
and the circular scale reads 16 divisions. Find the thickness of the glass plate. Zero of linear scale is not hidden
from circular scale when A and B touches each other.
Solution
least count = 0.01 mm
Also, we can say that,
Least count
Reading = Sleeve Reading + Thimble Reading
02_Measurements, General Physics_Part 1.indd 36
40
35
30
So, in general the correct measurement taken by
Screw Gauge is
Zero ⎞
⎛ Correct ⎞ ⎛ Observed ⎞ ⎛
⎜⎝ Length ⎟⎠ = ⎜⎝ Length ⎟⎠ + ⎜⎝ Correction ⎟⎠
Correct Length = Observed Length − Zero Error
45
sleeve is marked in divisions of 0.5 mm
Here also, we have
OR
5
L.C. =
pitch
number of divisions on circular scale
11/28/2019 6:51:35 PM
Chapter 2: Measurements and General Physics
LC =
1
mm = 0.01 mm
100
As zero is not hidden from circular scale when A
and B touches each other. Hence, the screw gauge
has positive error.
e = + n ( L.C. ) = 32 × 0.01 = 0.32 mm
Linear scale reading = 4 × ( 1 mm ) = 4 mm
Circular scale reading = 16 × ( 0.01 mm ) = 0.16 mm
⇒ Measured reading = ( 4 + 0.16 ) mm = 4.16 mm
⇒ Absolute reading = Measured reading −e
⇒ Absolute reading = ( 4.16 − 0.32 ) mm = 3.84 mm
Therefore, thickness of the glass plate is 3.84 mm
illustration 20
2.37
6 divisions below the reference line. When a wire is
placed between the jaws, 2 linear scale divisions are
clearly visible while 62 divisions on circular scale
coincide with the reference line. Determine the diameter of the wire.
solution
p 1 mm
=
= 0.01 mm
N
100
The instrument has a positive zero error.
L.C. =
(
)
(
)
e = + n L.C. = + 6 × 0.01 = + 0.06 mm
Linear scale reading = 2 × ( 1 mm ) = 2 mm
Circular scale reading = 62 × ( 0.01 mm ) = 0.62 mm
So, measured reading = 2 + 0.62 = 2.62 mm
i.e., True reading = 2.62 − 0.06 = 2.56 mm
The pitch of a screw gauge is 1 mm and there are 100
divisions on its circular scale. When nothing is put
in between its jaws, the zero of the circular scale lies
Test Your Concepts-IV
based on errors, significant figures, Vernier calliper and screw gauge
(Solutions on page H.9)
1. In a circuit the potential difference across a resistance is V = ( 8 ± 0.5 ) volt and the current in the
circuit is I = ( 2 ± 0.2 ) ampere . Calculate the resistance with absolute and the percentage error.
2. In an experiment for determining the density d of a
rectangular block of metal the length, breadth and
height are 5.12 cm , 2.56 cm , 0.37 cm and the
mass is 39.3 g . Calculate the maximum permissible absolute and relative error in the determination of density.
3. In a vernier callipers N divisions of vernier coincide with ( N − 1) divisions of main scale in which
length of 1 division is 1 mm. What is the least count
of the instrument in cm?
4. The length and the radius of a cylinder measured
with slide callipers are found to be 4.54 cm and
1.75 cm respectively. Calculate the volume of the
cylinder.
02_Measurements, General Physics_Part 1.indd 37
5. In the table shown, Column A gives few measured
values and Column B has the number of significant
figures to be correctly against the measured respective value. Complete the entries of Column B.
Column A
Column B
17236
________
510 m
________
270
________
4.20
________
7042.6
________
0.017
________
14
6.1 × 10
________
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2.38 JEE Advanced Physics: Mechanics – I
6. Use the rules studied for rounding off to give the
rounded off values to three significant figures.
Measured
Value
Rounded off value to Three
Significant Figures
7.364
________
8.3251
________
9.445
________
15.75
________
7.367
________
9.4450
________
15.7500
________
7. Calculate :
(a) 6.789 + 3.45 + 1.2 = ___________
(b) 12.63 − 10.2 = _____________
(c) 36.72 × 1.2 = ______________
(d)
1100 ms −1
10.2 ms −1
02_Measurements, General Physics_Part 1.indd 38
= ______________
8. The smallest division on main scale of a vernier
callipers is 1 mm and 10 vernier divisions coincide
with 9 scale divisions. While measuring the length
of a line, the zero mark of the vernier scale lies
between 10.2 cm and 10.3 cm and the third division of vernier scale coincide with a main scale
division.
(a) Determine the least count of the callipers
(b) Find the length of the line.
9. The pitch of a screw gauge is 1 mm and there are
100 divisions on the circular scale. In measuring
the diameter of a sphere there are six divisions on
the linear scale and forty divisions on circular scale
coincides with the reference line. Find the diameter
of the sphere.
10. In the diagram shown in figure, 9 MSD coincide
with 10 VSD. Find the magnitude and nature of
zero error.
0
0.5
0
5
1
10
Main Scale in cm
Vernier Scale
11/28/2019 6:51:40 PM
Chapter 2: Measurements and General Physics
2.39
Solved Problems
⇒
Problem 1
th
P.A.M. Dirac, a great Physicist of the 20 century
found that from the following basic constants a number having dimensional formula same as that of time
can be constructed i.e.,
(a) the charge on the electron e
(b) permittivity of free space ε o
(c) mass of the electron me
(d) mass of the proton mP
(e) speed of light c
(f) Universal Gravitational constant G.
Solution
According to the statement of the problem, we have
q
t ∝ e p ε 0 me−2 mp−1c r G s
Taking constant of proportionality as 1 (given in
question) and dimensions on both sides, we get
p
−2
−1
T = ( AT ) ( M −1 L−3 A 2 T 4 ) ( M ) ( M ) ×
q
( LT −1 )r ( M −1L3 T −2 )s
⇒
So, p = 4 , q = −2 , s = −1 , r = −3
Hence, we get
t = e 4 ε 0−2 me−2 mp−1c −3G −1
⇒
Obtain the relation for this Dirac’s number. You are
given that the desired number is proportional to mp−1
and me−2 . What is the significance of this number?
Assume constant of proportionality to be 1.
T = M − q − 3 − s L−3 q + r + 3 s T p + 4 q − r − 2 s A p + 2 q −q − 3 − s = 0 …(1)
−3 q + r + 3 s = 0 …(2)
p + 4 q − r − 2s = 1 …(3)
p + 2q = 0 …(4)
−2q = 4 02_Measurements, General Physics_Part 1.indd 39
…(7)
18
11
t = 3.36 × 10 s ≅ 10 years
Dirac estimated this time to be the age of universe.
Problem 2
The mean life of the neutral elementary particle pion
is 2 × 10 −7 ns . The age of the universe is about 4 × 109
year. Find a time that is midway between these two
times on the logarithmic scale.
Solution
log e t =
log e t1 + log e t2
2
1
log e ( t1t2 ) = log e t1t2
2
Taking antilog both sides, we get
⇒
log e t =
t = t1t2
…(5)
t1 = 2 × 10 −7 ns = 2 × 10 −16 s
t2 = 4 × 109 year = 4 × 109 × 365 × 24 × 60 × 60 s
q − 2q + s = 1
Adding (1) and (6), we get
Now substituting the values of all constants in equation (7), we get the value of this time t as
From (5) p = −2q , put in (5) to get
−q + s = 1 e
ε 02 me2 mpGc 3
Also, you can verify the dimensional correctness of
equation (7) by substituting the dimensional formulae
of the respective quantities on the right.
Adding (2) and (3) for eliminating r , we get
⇒
t=
4
Let t be the time half way between these two on the
logarithmic scale. Then
Applying Principle of Homogeneity, we get
q+ p+ s = 1
q = −2 …(6)
⇒
t2 = 1.26 × 1017 s So, t1t2 ≅ 25
⇒
t1t2 = 5 s
11/28/2019 6:51:47 PM
2.40 JEE Advanced Physics: Mechanics – I
Problem 3
Problem 4
Derive by the method of dimensions, an expression
for the volume of a liquid flowing out per second
through a narrow pipe. Assume that the rate of flow
of liquid depends on
Finding dimensions of resistance R and inductance
⎛ L⎞
L , speculate what physical quantities ⎜ ⎟ and
⎝ R⎠
⎛ 1⎞ 2
⎜⎝ ⎟⎠ Li represents ?
2
(a) the coefficient of viscosity η of the liquid
(b) the radius r of the pipe and
Solution
⎛ p⎞
(c) the pressure gradient ⎜ ⎟ along the pipe.
⎝ l⎠
π
Take k =
8
As e = L
So
Solution
Let volume flowing out per second through the pipe
be given by
a b ⎛ p⎞
c
V = kη r ⎜ ⎟ ⎝ l⎠
…(1)
where k = a dimensionless constant. Dimensions of
the various quantities are
volume L3
=
= L3 T −1
[V ] =
time
T
[ η ] = ML T
−1
−1
, [r] = L
−2
⎡ p ⎤ MLT
⎢⎣ l ⎥⎦ = L2 .L
= ML−2 T −2
Substituting these dimensions in equation (1), we get
a
⎡ L3 T −1 ⎤ = ⎡ ML−1T −1 ⎤ [ L ]b ⎡ ML−2 T −2 ⎤
⎣
⎦ ⎣
⎦
⎣
⎦
⇒
c
M 0 L3 T −1 = M a + c L− a + b − 2c T − a − 2c Equating the powers of M , L and T , we get
a + c = 0 , − a + b − 2c = 3 , − a − 2c = −1
On solving, a = −1 , b = 4 , c = 1 .
−1 4 ⎛ p ⎞
⇒
V = kη r ⎜ ⎟
⎝ l⎠
⇒
πr4 p
V=
8ηl
1
02_Measurements, General Physics_Part 1.indd 40
⎡ W ⎤ ⎡ t ⎤ ML2T −2 T
×
⎥⎢ ⎥ =
AT
A
⎣ q ⎦⎣i ⎦
[L] = ⎢
i.e., [ L ] = ML2 T −2 A −2
and as V = IR, i.e., R =
V
I
⎡ W ⎤ ML2 T −2
⎥=
ATA
⎣ qA ⎦
So
[R] = ⎢
⇒
[ R ] = ML2T −3 A −2 So
2 −2 −2
1
⎡ L ⎤ ML T A
⎢⎣ R ⎥⎦ = ML2 T −3 A −2 = T −1 = T
⎤
⎡1
and ⎢ Li 2 ⎥ = ( ML2 T −2 A −2 ) ( A 2 ) = ML2 T −2
⎣2
⎦
force
⎡ p ⎤ pressure
⎢ l ⎥ = length = area × length
⎣ ⎦
⇒
di
dt
i.e., L = e
dt
di
⎛ L⎞
Now as ⎜ ⎟ has dimensions of time and hence is
⎝ R⎠
⎛ 1⎞
called time constant of L − R circuit and ⎜ ⎟ Li 2 has
⎝ 2⎠
dimensions of work or energy, so it represents magnetic energy stored in a coil.
Problem 5
The Renold’s number nR for a liquid through a pipe
depends upon the density of the liquid d , the coefficient of viscosity η , the speed of the liquid v and the
radius of the tube r . Obtain by dimensional analysis
an expression for nR . Given nR ∝ r .
Solution
As per the statement of the question, we get
[Poiseuille’s equation]
nR ∝ r …(1)
11/28/2019 6:51:55 PM
2.41
Chapter 2: Measurements and General Physics
nR ∝ d a …(2)
b
nR ∝ η …(3)
nR ∝ vc …(4)
Combining (1), (2), (3) and (4), we get
⇒
ΔRS
× 100 = 6.25%
RS
⇒
RS = 16 Ω ± 6.25% Similarly
nR ∝ rd aηb vc
⇒
a b c
nR = krd η v M L T = L ( ML
⇒
0
1
1
1
=
+
RP R1 R2
…(5)
where k is a dimensionless constant
Taking dimensions on both sides of (5), we get
0 0
) ( ML T ) ( LT )
−3 a
−1
−1 b
−1 c
M 0 L0 T 0 = M a + b L−3 a − b + c +1T − b − c So, from Principle of Homogeneity, we get
⇒
⇒
⇒
a = −b and c = −b So, we get
−3 ( −b ) − b − b + 1 = 0
⇒
3b − 2b + 1 = 0 ⇒
b = −1 Hence a = 1 , b = −1 and c = 1.
So, we get the final relation as
nR =
kvrd
η
Two resistances are expressed as R1 = ( 4 ± 0.5 ) Ω
and R2 = ( 12 ± 0.5 ) Ω . Calculate the net resistance
when both are connected in series and in parallel
with absolute and %age error.
RS = R1 + R2 = 16 Ω
ΔRS = ΔR1 + ΔR2 = 1 Ω
02_Measurements, General Physics_Part 1.indd 41
(
)
(
ΔRP
=
RP2
ΔR1
+
2
R1
ΔR2
)
R22
where
R1 R2
= 3Ω
R1 + R2
ΔRP ⎛ 0.5 0.5 ⎞
=⎜
+
( 3)
⎝ 16 144 ⎟⎠
RP
⇒
ΔRP
15
=
RP
144
⇒
% RE = 10.42%
Problem 7
In an experiment the following readings were
recorded L = 2.890 m , M = 3.00 kg , l = 0.87 cm and
and using the
MgL
formula for Young’s modulus Y =
, calculate
π r 2l
the maximum permissible error in the value of Y .
Solution
Y=
⇒
R1 R2
RR
= 1 2 =3Ω
R1 + R2
RS
ΔRS
1
× 100 =
× 100%
RS
16
)
⇒
Solution
⇒
(
D = 0.041 cm . Taking g = 9.8 ms −2
Problem 6
RP =
Δ RP−1 = Δ R1−1 + Δ R2−1
RP =
−3 a − b + c + 1 = 0 ,
−b − c = 0
RP−1 = R1−1 + R2−1
⇒ ( −1)RP−2 ΔRP = ( −1)R1−2 ΔR1 + ( −1)R2−2 ΔR2
a+b = 0,
⇒
⎛ 4 g ⎞ ML
=⎜
⎟
π r l ⎝ π ⎠ D2 l
MgL
2
{
∵ r=
D
2
}
ΔY ΔM ΔL
ΔD Δl
=
+
+2
+
Y
M
L
D
l
Since maximum permissible error can be equal to the
least count of the device taking the measurement, so
we have
ΔM = 0.01 kg , ΔL = 0.001 m ,
ΔD = 0.001 cm and Δl = 0.01 cm
11/28/2019 6:52:05 PM
2.42 JEE Advanced Physics: Mechanics – I
⇒
%
ΔY ⎡ 0.01 0.001 ⎛ 0.001 ⎞ ⎛ 0.01 ⎞ ⎤
=
+
+ 2⎜
+
× 100
⎝ 0.041 ⎟⎠ ⎜⎝ 0.87 ⎟⎠ ⎥⎦
Y ⎢⎣ 3
2.89
⇒
%
ΔY 1
0.1
2 ( 0.1 )
1
= +
+
+
Y
3 2.890 0.041 0.87
⇒
%
ΔY 1 10 200 100
= +
+
+
Y
3 289 41
87
⇒
%
ΔY
= 0.34 + 0.035 + 4.9 + 1.2
Y
⇒
%
ΔY
= 6.5%
Y
Problem 8
N divisions on the main scale of a vernier callipers
coincide with N + 1 divisions on the vernier scale. If
each division on the main scale is of a units, determine the least count of the instrument.
02_Measurements, General Physics_Part 1.indd 42
Solution
( N + 1 ) divisions on the vernier scale = N divisions
on main scale
N
1 division on vernier scale =
divisions on main
N +1
scale
Each division on the main scale is of a units.
⎛ N ⎞
So, 1 division on vernier scale = ⎜
a unit = a ′
⎝ N + 1 ⎟⎠
(say)
Least count = 1 MSD − 1 VSD
a
⎛ N ⎞
⇒ Least count = a − a ′ = a − ⎜
a=
⎝ N + 1 ⎟⎠
N +1
11/28/2019 6:52:08 PM
Chapter 2: Measurements and General Physics
2.43
Practice Exercises
Single Correct Choice Type Questions
This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
The amount of heat produced in an electric circuit
depends upon the current ( I ) , resistance ( R ) and
time ( t ) . If the errors created in the measurements of
the above quantities are 2%, 1% and 1% respectively
then the maximum possible error will be
(A) 1%
(B) 2%
(C) 3%
(D) 6%
2.
The ratio of MKS and CGS units of the coefficient of
viscosity is
(A) 10
(B) 10 2
(C) 10 −1
3.
4.
5.
The unit of the reduction factor of tangent galvanometer is
(A) second
(B) ampere
(C) radian
(D) tesla
Which of the following pairs have the same
dimensions?
(A) Pressure, stress and force
(B) Work, power and power factor
(C) Impulse and momentum
(D) Force, impulse and momentum
The unit of coefficient of viscosity in MKS system is
(B) kgm -1s -1
(A) mkg -1s -1
(C) m -1s -1
6.
7.
(D) mkgs
The unit of Vander Waal’s gas constant a in MKS
­system will be
(A) kgm 5s −1
(B)
(C) kgm 5s −2
(D) kgm 5s
The dimensions of
sions of
(A) velocity
(C) time
8.
(D) 10 −2
kgm 4s −2
1
are equivalent to the dimenμε
(B) acceleration
(D) energy
How many significant figures are there in the numbers
108.023 and 0.19?
(A) 6 and 2
(B) 2 and 6
(C) 3 and 5
(D) 5 and 3
02_Measurements, General Physics_Part 2.indd 43
9.
The number of particles crossing the unit area normal
to x-axis per second is represented by the following
n − n1
formula n = D 2
, where D is the coefficient of
x2 − x1
diffusion and n1 and n2 are the number of molecules
per unit volume at x1 and x2 , respectively, then the
dimensions of D are
(A) M 0 L2T −1
(C) M 0 LT −2
(B) M 0 L2T −2
(D) M 0 LT −1
L
10. The dimensions of
and RC are equivalent to the
R
dimensions of
(A) time
(B) frequency
(C) wavelength
(D) energy
11. In the formula V = d a Eb , if V , E and d are the velocity of longitudinal waves, bulk modulus of elasticity
and density of the gaseous medium respectively, then
the values of a and b are respectively.
1
1
and
2
2
(B)
1
1
and −
2
2
1
1
and
2
2
(D)
1
1
and
2
2
(A) −
(C)
1 q2
1
1
12. The dimensional formula for CV 2 or
or qV
2C
2
2
is
−2
2 −2
(B) ML T
(A) MLT
2 2
(C) ML T
(D) ML−2T −3
13. If force ( F ) , length ( L ) and time ( T ) are presumed
to be the fundamental units, then the dimensional
­formula of mass will be
(A) FL−1T 2
(B)
FL−1T −2
(C) FL−1T −1
(D) FL2T 2
14. If L, C and R denote the inductance, capacitance and
resistance respectively, the dimensional formula for
C 2 LR is
(A) ML−2T −1
(B)
(C) M −1L−2T 6 A 2
(D) M 0 L0T 2
M 0 L0T 3
11/28/2019 6:49:23 PM
2.44 JEE Advanced Physics: Mechanics – I
15. The position of a particle at time t is given by the relav
tion xt = ⎛⎜ 0 ⎞⎟ ( 1 − c −α t ) , where v0 is a constant and
⎝ α⎠
α > 0 . The dimensions of v0 and α are respectively
(A) M 0 L1T −1 and T −1
(B)
M 0 L1T 0 and T −1
(C) M 0 L1T −11
(D) M 0 L1T −1 and T
A x
, where A and B
x2 + B
are dimensional constants then dimensional formula
for AB is
x from a fixed origin as U =
7
9
(C) M 2 L2 T −2
11
(B)
ML 2 T −2
13
(D) ML 2 T −3
17. Using dimensional analysis you can check on some
results. In the integral
the value of n is
(A) 1
(C) 0
18. The
dx
−1 ⎛ x
⎞
∫ ( 2ax − x ) = a sin ⎜⎝ a − 1⎟⎠
n
2 12
(B)
−1
1
2
stationary
(D)
equation of the
wave is
⎛ 2π ct ⎞
⎛ 2π x ⎞ . Which of the following
y = 2 a sin ⎜
cos ⎜
⎝ λ ⎟⎠
⎝ λ ⎟⎠
statements is wrong?
(A) The unit of ct is same as that of λ
(B) The unit of x is same as that of λ
2π c
2π x
(C) The unit of
is same as that of
λ
λt
c
x
(D) The unit of
is same as that of
λ
λ
19. A physical quantity is measured and its value is found
to be nu where n is numerical value and u is unit.
Then which of the following relations is true
(A) n ∝ u2
(B)
(C) n ∝ u
(D) n ∝
n∝u
1
u
20. In C.G.S. system the magnitude of the force is 100 dyne.
In another system where the fundamental physical
quantities are kilogram, metre and minute, the magnitude of the force is
(A) 0.036
(B) 0.36
(C) 3.6
(D) 36
21. The temperature of a body on Kelvin scale is found
to be X K . When it is measured by a Fahrenheit thermometer, it is found to be X °F . Then X is
02_Measurements, General Physics_Part 2.indd 44
(B) 574.25
(D) 40
22. Which relation is incorrect?
(A) 1 calorie = 4.18 joule (B) 1 Å = 10 −10 m
(C) 1 MeV = 1.6 × 10 −13 J
16. The potential energy of a particle varies with distance
(A) ML2 T −2
(A) 301.25
(C) 313
(D) 1 newton = 10 −6 dyne
23. To determine the Young’s modulus of a wire, the forF L
mula is Y =
, where L is the length, A is the
A ΔL
area of cross-section of the wire, ΔL is the change
in length of the wire when stretched with a force F .
The conversion factor to change it from CGS to MKS
­system is
(A) 1
(B) 10
(C) 0.1
(D) 0.01
24. Conversion of 1 MW power on a new system having
basic units of mass, length and time as 10 kg , 1 dm
and 1 minute respectively is
(A) 2.16 × 1012 unit
(C) 2.16 × 1010 unit
(B) 1.26 × 1012 unit
(D) 2 × 1014 unit
25. In two systems, the relations among velocity, acceleraα2
tion and force are respectively v2 =
v1 , a2 = αβ a1
β
F1
. If α and β are constants then mass,
and F2 =
αβ
length and time in two systems are related to each as
(A) M2 =
(B)
M2 =
α
α 3T1
α2
M1 , L2 = 2 L1 , T2 =
β
β
β
1
α3
α
M1 , L2 = 3 L1 , T2 = T1 2
α β
β
β
2 2
(C) M2 =
α
α3
α2
M1 , L2 = 2 L1 , T2 = T1
3
β
β
β
(D) M2 =
α2
α
α3
M
L
=
L
T
=
T1
,
,
1
2
1
2
β2
β2
β3
26. If the present units of length, time and mass ( m, s, kg )
1
are changed to 100 m, 100 s, and
kg, then
10
(A)the new unit of velocity is increased 10 times
1
(B)the new unit of force is decreased
times
1000
(C) the new unit of energy is increased 10 times
(D)the new unit of pressure is increased 1000 times
27. Suppose we employ a system in which the unit of
mass equals 100 kg, the unit of length equals 1 km
and the unit of time 100 s and call the unit of energy
eluoj (joule written in reverse order), then
11/28/2019 6:49:33 PM
2.45
Chapter 2: Measurements and General Physics
(A) 1 eluoj = 10 4 joule
(B) 1 eluoj = 10 −3 joule
(C) 1 eluoj = 10 −4 joule
(D) 1 joule = 10 3 eluoj
28. If 1 gcms −1 = x Ns , then number x is equivalent to
(A) 1 × 10 −1
(B)
3 × 10 −2
(C) 6 × 10 −4
(D) 1 × 10 −5
29. A highly rigid cubical block A of small mass M and
side L is fixed rigidly onto another cubical block B of
the same dimensions and of low modulus of rigidity
η such that the lower face of A completely covers the
upper face of B . The lower face of B is rigidly held
on a horizontal surface. A small force F is applied
perpendicular to one of the side faces of A . After the
force is withdrawn block A executes small oscillations. The time period of which is given by
(A) 2π
Mη
L
(B)
2π
L
Mη
(C) 2π
ML
η
(D) 2π
M
ηL
1
1
1
1
(B)
1
1
−
1
(C) c 2 G 2 h 2
−
1
1
c 2G 2 h 2
−
1
1
1
(D) c 2 G 2 h 2
31. If P represents radiation pressure, C represents speed
of light and Q represents radiation energy striking a
unit area per second, the non-zero integers x , y and
z such that P xQ y C z is dimensionless, are
(A) x = 1 , y = 1 , z = −1
(C) x = −1 , y = 1 , z = 1
(B)
x = 1 , y = −1 , z = 1
(D) x = 1 , y = 1 , z = 1
32. The volume V of water passing through a point of a
uniform tube during t seconds is related to the crosssectional area A of the tube and velocity u of water
by the relation V ∝ Aα uβ tγ , which one of the following will be true
(A) α = β = γ
(B) α ≠ β = γ
(C) α = β ≠ γ
(D) α ≠ β ≠ γ
33. Given that the amplitude A of scattered light is
(i)directly proportional to the amplitude ( A0 ) of
incident light
02_Measurements, General Physics_Part 2.indd 45
(C) A ∝
1
λ3
(D) A ∝
1
λ4
34. Each side a cube is measured to be 7.203 m. The volume of the cube up to appropriate significant figures is
(A) 373.714
(B) 373.71
(C) 373.7
(D) 373
35. Candela is the unit of
(A) acoustic intensity
(C) luminous intensity
(B) electric intensity
(D) magnetic intensity
36. The time taken by sunlight to penetrate a window
pane is of the order of
(A) 10 −5 s
(C) 10 −11 s
30. If the velocity of light ( c ) , gravitational constant ( G )
and Planck’s constant ( h ) are chosen as fundamental
units, then the dimensions of mass in new system is
(A) c 2 G 2 h 2
(ii)
directly proportional to the volume (V) of the
scattered particle
(iii)inversely proportional to the distance ( r ) from
the scattered particle
(iv)depends upon the wavelength ( λ ) of the scattered light then
1
1
(B) A ∝ 2
(A) A ∝
λ
λ
(B) 10 −7 s
(D) 10 −19 s
37. Which of the following is the unit of latent heat?
(A) J
(B) J mol −1
(C) J kg −1
(D) J kg −1mol −1
38. The energy E radiated per unit area per second by a
black body at temperature T is given by E = σ T 4 , where
s is the Stefan’s constant. The dimensions of s are
(A) MT 2 K −2
(B) MT −3 K −4
(C) MT 3 K −4
(D) ML4 T −3 K −4
39. If length ( L ) , mass ( M ) and force ( F ) are taken as
fundamental quantities, then the dimensions of time
will be
1
1
−
1
−
1
1
(A) M 2 L2 F 2
(B)
(C) M 2 L2 T −2
(D) MLF 2
1
M 2 L2 F 2
1
40. The surface tension of a liquid is 70 dynecm −1 . It may
be expressed in MKS as
(B) 7 × 10 −2 Nm −1
(A) 70 Nm −1
(C) 7 × 10 2 Nm −1
(D) 7 × 10 3 Nm −1
41. Temperature can be expressed as a derived quantity in
terms of any of the following
(A) length and mass
(B) mass and time
(C) length, mass and time
(D) in terms of none of these
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2.46 JEE Advanced Physics: Mechanics – I
42. The density of a material is 8 gcc −1 . In a system in
which the unit of length is 5 cm and unit of mass is
20 g , the density of material will be
(A) 40
(B) 50
(C) 24
(D) 80
43.
erg m −1 can be the unit of the measure of
(A) force
(C) power
(B) momentum
(D) acceleration
44. Which one of the following has the dimensions of
pressure?
(A)
(C)
M
L2T 2
ML
T2
(B)
M
LT
M
(D)
LT 2
45. In a particular system, the units of length, mass and
time are chosen to be 10 cm, 10 g and 0.1 s respectively.
The unit of force in this system will be equivalent to
(A) 0.1 N
(B) 1 N
(C) 10 N
(D) 100 N
46. The dimensions of resistivity in terms of M , L , T
and Q are
(A) ML3 T −1 Q −2
(C) ML2 T −1 Q −1
(B) ML3 T −2 Q −1
(D) MLT Q −1
47. The SI unit of pressure is
(A) dyne cm −2
(B) atmosphere
(D) cm of Hg
(C) pascal
48. A certain physical quantity is calculated from the
π( 2
­formula
a − b 2 ) h , where h , a and b are all
3
lengths. The quantity being calculated is
(A) Velocity
(B) Length
(C) Area
(D) Volume
49. Joulesecond is a unit of
(A) energy
(C) power
(B) momentum
(D) angular momentum
50. If x = at + bt 2 + c , where x is in metre and t in hour
(hr), then the unit of a , b and c are
(A) m , mhr −1 , mhr −2
(B) mhr −1 , mhr −2 , m
(C) m 2 hr −1 , mhr −1 , m
51. The dimensions of
(A) velocity
(C) capacitance
(D) mhr −2 , mhr −1 , m
1
are those of
∈0 μ0
(B) time
(D) distance
52. Taking into account the significant figures, the product
of 109.832 and 0.6107 should be written as
02_Measurements, General Physics_Part 2.indd 46
(A) 67.0744
(C) 67.07
(B) 67.1
(D) 67.074402
53. The dimensions of torque are
(A) ML2 T −2
(B)
(C) MLT −1
MLT 2
(D) MT −2
54. Which of the following quantities has the dimensional
formula ML2 T −3 ?
(A) Bulk modulus
(B) Coefficient of viscosity
(C) Energy
(D) Power
55. The dimensions of the ratio of angular to linear
momentum is
(B) M 1 L1 T 1
(A) M 0 L1T 0
(C) M 1 L2 T −1
(D) M −1 L−1 T −1
56. The random error in the arithmetic mean of 100 observations is x . The random error in the arithmetic mean
of 400 observations would be
x
(A) 4x
(B)
4
x
(C) 2x
(D)
2
57. The velocity v of a particle is given in terms of time
b
t by the equation v = at +
. The dimensions of
t+c
a, b , c are respectively
(A) L2 , T , LT 2
(C) LT −2 , L , T
(B) LT 2 , LT , L
(D) L , LT , T 2
58. The dimensions of mc 2 are
(B) ML2T −1
(A) MLT −1
2 −2
(C) ML T
(D) ML2T 2
59. Which of the following is not a unit of energy?
(A) Ws
(B) kgms −1
(C) Nm
(D) joule
60. 1 micron ( μ ) is
(A) 10 −9 m
(C) 10 −6 m
(B) 10 −12 m
(D) 10 −15 m
61. The unit of Stefan’s constant σ is
(A)
watt 4
mK 4
(B)
calorie
m 2K 4
(C)
watt
m 2K 4
(D)
joule
m 2K 4
62. Newtonsecond is a unit of
(A) force
(B) impulse
(C) momentum
(D) energy
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Chapter 2: Measurements and General Physics
63. If DX is absolute error in the measurement of X and
ΔY is absolute error in the measurement of Y , then
maximum absolute error in the measurement of difference of X and Y is
(A) ± ( ΔX − ΔY )
(B) ( ΔX + ΔY )
(C) ±
ΔX
ΔY
(D) ±ΔX ΔY
64. The dimensional formula of universal gravitational
constant is
(B) ML3T −2
(A) MLT −2
(C) M −1L3T −2
(D) MLT 2
65. If the units of M and L are doubled then the unit of
kinetic energy will become
(A) 8 time
(B) 16 time
(C) 4 time
(D) 2 time
66. Dimensional analysis gives
(A)no information about dimensionless constants
(B) information about dimensionless constants
(C)
information about dimensionless constants if
quantity does not depend upon more than three
variables
(D)
information about dimensionless constants if
quantity depends upon single variable
67. If force ( F ) , area ( A ) and density ( D ) are taken as
fundamental quantities, the dimensional formula for
Young’s modulus will be
(B) FA −1 D0
(A) FA −2 D2
−1 −1 −1
(C) F A D
(D) FA −1 D
68. The dimensions of the coefficient of viscosity are
[ ML−1T −1 ] . To convert the CGS unit poise ( P ) to the
MKS unit poiseuille ( PI ) , the poise has to be multiplied by
(A) 10 −1
(C) 109
(B) 10
(D) 107
69. The dimensions of “light-year” are
(A) LT −1
(B) T
(C) ML2 T −2
(D) L
70. The velocity of a particle is given by v = at 2 + bt + c .
If v is measured in ms −1 and t is measured in s , the
unit of
(A) a is ms −1
(B) b is ms −1
(C) c is ms −1
(D) a and b are same but that of c is different
02_Measurements, General Physics_Part 2.indd 47
2.47
71. The pair having the same dimension is
(A) angular momentum, work
(B) work, torque
(C) potential energy, linear momentum
(D) kinetic energy, velocity
t
72. Given that y = a cos ⎛ − qx ⎞ , where t represents
⎜⎝ p
⎟⎠
time in second and x distance in metre. Which of the
following statements is true?
(A) The unit of x is same as that of q.
(B) The unit of x is same as that of p.
(C) The unit of t is same as that of q.
(D) The unit of t is same as that of p.
73. Electronvolt ( eV ) is a unit of
(A) charge
(B) potential
(C) energy
(D) coulomb repulsion
74. A system has basic dimensions as density [ D ] , velocity [ V ] and area [ A ] . The dimensional representation of force in this system is
[ AV 2D ]
(C) [ AVD2 ]
(A)
[ A2VD ]
(D) [ A0VD ]
(B)
ΔV
, where ε 0 is the perΔt
mittivity of free space, L is a length, ΔV is a potential
difference and Δt is a time interval. The dimensional
formula for X is the same as that of
(A) resistance
(B) charge
(C) voltage
(D) current
75. A quantity X is given by ε 0 L
L
during growth
R
and decay of current in an inductive circuit are same
as those of
(A) constant
(B) resistance
(C) current
(D) time
76. The dimensions of time constant
77. If the speed of light ( c ) acceleration due to gravity
( g ) and pressure ( p ) are taken as the fundamental
quantities, then the dimensions of length will be
(A) c 2 g −1 p0
(B)
cg 0 p −1
(C) c −1 gp0
(D) cg −1 p0
a
and Δa and Δb are the errors in the meab
surement of a and b , respectively, then the maximum
percentage error in the value of x will be
78. If x =
11/28/2019 6:50:07 PM
2.48 JEE Advanced Physics: Mechanics – I
Δa Δb ⎞
(A) ⎛⎜
+
⎟ × 100
⎝ a
b ⎠
(B)
⎛ Δa Δb ⎞
−
⎜⎝
⎟ × 100
a
b ⎠
Δa
Δb ⎞
(C) ⎛⎜
+
× 100
⎝ a − b a − b ⎟⎠
Δa
Δb ⎞
(D) ⎛⎜
−
× 100
⎝ a − b a − b ⎟⎠
86. If r is the mean density of the earth, r its radius, g the
acceleration due to gravity and G the gravitational
constant, on the basis of dimensional consistency, the
correct expression is
3G
4π rg
3g
(C) ρ =
4π rG
(A) ρ =
79. If the orbital velocity of a planet is given by
v = G a M b Rc , then
(A) a =
(B)
1
1
1
, b= , c=−
3
3
3
1
1
1
, b=− , c=
2
2
2
1
1
1
(D) a = , b = − , c = −
2
2
2
(C) a =
80. In an experiment to determine acceleration due to
gravity by simple pendulum, a student commits 1%
positive error in the measurement of length and 3%
negative error in the measurement of time period. The
percentage error in the value of g will be
(A) 7%
(B) 10%
(C) 4%
(D) 3%
81. The best method to reduce random errors is
(A)to change the instrument used for measurement
(B) to take help of experienced observer
(C)to repeat the experiment many times and to take
the average results
(D) None of these
82. The least count of an instrument is 0.01 cm. Taking all
precautions the most possible error in the measurement can be
(A) 0.005 cm
(B) 0.001 cm
(C) 0.01 cm
(D) 0.02 cm
83. The fundamental frequency of vibration of a stretched
1 T
, where the symbols have
2l m
their usual meanings. The dimensions of m are
string is given by ν =
(B) ML0 T 0
(D) M −1 LT 0
84. Which of the following pairs has the same dimensions?
(A) Pressure and density
(B) Impulse and momentum
(C) Stress and strain
(D) Momentum and inertia
85. The dimensions of PV are equivalent to those of
(A) work
(B) force
(C) pressure
(D) volume
02_Measurements, General Physics_Part 2.indd 48
rG
12π g
(D) ρ =
3r
4π gG
1
87. The dimension of ⎛⎜ ⎞⎟ ε 0E2 ( ε 0 : permittivity of free
⎝ 2⎠
space; E : electric field) is
(A) MLT −1
(B) ML−1T −2
−2
(C) MLT
(D) ML2T −1
1
1
1
a= , b= , c=−
2
2
2
(A) ML−1 T 0
(C) ML−1 T
ρ=
(B)
88. Of the following quantities, which one has dimensions
different from the remaining three?
(A) Energy per unit volume
(B) Force per unit area
(C)Product of voltage and charge per unit volume
(D) Angular momentum per unit mass
89. The time dependence of a physical quantity P is
given by P = P0 exp ( −α t 2 ) where α is a constant and
t is time. The constant α
(A) is dimensionless
(B) has dimensions of T −2
(C) has dimensions as that of P
(D)has dimensions equal to the dimensions of PT −2
90. The dimensional formula of magnetic moment is
(A) M −2 A
(C) L2 A
(B) M 2 A
(D) L−2 A
91. If energy E , velocity V and time T are taken as the
fundamental units, the dimensional formula for surface tension is
(B) E −2VT −2
(A) EV −2T −2
(C) E −2V −2T
(D) E −2V −2T −2
92. In the formula X = 3YZ 2 , X and Z have dimensions
of capacitance and magnetic induction respectively.
What are the dimensions of Y in MKSQ system?
93.
(A) M −3 L−1T 3Q 4
(B)
M −3 L−2T 4Q 4
(C) M −2 L−2T 4Q 4
(D) M −3 L−2T 4Q1
E, m, L, G denote energy, mass, angular momentum
and gravitational constant respectively. The dimenEL2
sions of 5 2 will be that of
mG
(A) angle
(B) length
(C) mass
(D) time
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Chapter 2: Measurements and General Physics
94. Out of the following expressions for the couple per
unit twist, time period of a compound pendulum, frequency of a stretched wire and time period of a simple
pendulum respectively, the dimensionally inconsistent one is
(A) C =
ηπ r 4
2l
(B)
f =
1 T
l ρ
(D) T = 2π
(C)
T = 2π
k2 l
+
g g
l
g
where η is coefficient of rigidity, r is the radius of the
wire, K is the radius of gyration, and ρ is mass per
unit length and l is the length corresponding to the
particular case.
95. In the radioactive decay law N = N 0 e
sions of λ are
− λt
, the dimen-
M 0 L0 T −1
(A) M 0 L0 T 0
(B)
(C) M 0 L0 T
(D) ML0 T −1
96. Planck’s constant has the dimensions of
(A) force
(B) energy
(C) linear momentum
(D) angular momentum
97. The SI unit of mobility of charges ( μ ) is
(A) Cskg
−1
(C) Cskg
(B)
Ckgs
−1
(D) Cs −1kg −1
98. Turpentine oil is flowing through a tube of length l
and radius r . The pressure difference between the
two ends of the tube is p ; the viscosity of the oil is
p ( r 2 − x2 )
where v is the velocity of oil
4vl
at a distance x from the axis of the tube. From this
relation, the dimensions of viscosity η are
given by η =
2.49
(A)It is both dimensionally as well as numerically
correct.
(B)It is neither dimensionally correct nor numerically correct.
(C)It is dimensionally correct but not numerically.
(D)It is numerically correct but not dimensionally.
101. If V denotes the potential difference across the plates
of a capacitor of capacitance C , the dimensions of
CV 2 are
(A) MLT −2
(B) M 2 LT −1
(C) ML2T −2
(D) not expressible in terms of M , L , T
102. Unit of amplification factor is
(A) ohm
(B) mho
(D) None of these
(C) AV −1
103. A physical quantity is represented by the relation
Y = M a LbT − c . If the percentage errors in the measurement of M , L and T are respectively α % , β % and
γ %, respectively then the total error will be
(B) ( α a − βb − γ c ) %
(A) ( α a + βb − γ c ) %
(C)
( α a + βb + γ c ) %
(D)
( α a − βb + γ c ) %
104. Given that T stands for time period and l stands for
the length of simple pendulum. If g is the acceleration due to gravity, then which of the following state⎛ l⎞
ments about the relation T 2 = ⎜ ⎟ is correct?
⎝ g⎠
(A)
It is correct both dimensionally as well as
numerically
(B)It is neither dimensionally correct nor numerically
(C)It is dimensionally correct but not numerically
(D)If is numerically correct but not dimensionally
1
is equivalent to that of
2π LC
(A) time period
(B) frequency
(C) wave length
(D) wave number
105. Select the incorrect statement(s).
(A)If the zero of vernier scale does not coincide with
the zero of the main scale, then the vernier calliper is said to be having zero error
(B)Zero correction has a magnitude equal to zero
error but sign is opposite to that of zero error
(C)Zero error is positive when the zero of vernier
scale lies to the left of the zero of the main scale
(D)Options (B) and (C) are incorrect.
v2
gives the angle of banking of the
rg
cyclist going round the curve. Here v is the speed of
cyclist, r is the radius of the curve and g is acceleration due to gravity. Which of the following statements
about the relation is true?
106. 1 cm on the main scale of a vernier calliper is divided
into 10 equal parts. If 10 divisions of vernier coincide
with 8 small divisions of main scale, then the least
count of the calliper is
(A) 0.01 cm
(B) 0.02 cm
(C) 0.05 cm
(D) 0.005 cm
MLT −1
(A) M 0 L0T 0
(B)
(C) ML2T −2
(D) ML−1T −1
99. The SI unit of
100. Given that tan θ =
02_Measurements, General Physics_Part 2.indd 49
11/28/2019 6:50:23 PM
2.50 JEE Advanced Physics: Mechanics – I
107. The vernier constant of a travelling microscope is
0.001 cm. If 49 main scale divisions coincide with 50
main scale divisions, then the value of 1 main scale
division is
(A) 0.1 mm
(B) 0.5 mm
(C) 0.4 mm
(D) 1 mm
108. Which of the following has the largest least count?
(A) Spherometer
(B) Vernier callipers
(C) Screw gauge
(D) Metre scale
109. 1 cm of main scale of a vernier caliper is divided
into 10 divisions. The least count of the callipers is
0.005 cm, then the vernier scale must have
(A) 10 divisions
(B) 20 divisions
(C) 25 divisions
(D) 50 divisions
1
cm. The
110. Least count of a vernier calliper is
10 N
value of one division on the main scale is 1 mm. Then
the number of divisions of the main scale that coincide with N divisions of vernier scale is
N
10
(A) 10 N
(B)
(C) ( N − 1 )
(D) N − 10
111. Each division on the main scale is 1 mm. Which of the
following vernier scales give vernier constant equal
to 0.01 mm?
(A) 9 mm divided into 10 divisions
(B) 90 mm divided into 100 divisions
(C) 99 mm divided into 100 divisions
(D) 9 mm divided into 100 divisions
112. The least count of vernier callipers is 0.01 cm. Then
the error in the measurement is
(A) > 0.01 cm
(B)
≥ 0.01 cm
(C) < 0.01 cm
(D) ≤ 0.01 cm
113. A person performs an experiment with vernier callipers and takes 100 readings. He repeats the same
experiment but now takes 400 readings. Then the
probable error is
(A) the same
(B) halved
(C) doubled
(D) reduced by one fourth
114. Vernier constant is the
(A)value of one MSD divided by total number of
divisions on the main scale
(B)value of one VSD divided by total number of
divisions on the vernier scale
02_Measurements, General Physics_Part 2.indd 50
(C)
total number of divisions on the main scale
divided by total number of divisions on the vernier scale
(D)difference between value of one main scale division and one vernier scale division
115. Which of the following uses angular vernier?
(A) Metre scale
(B) Vernier callipers
(C) Spherometer
(D) Both (A) and (B)
116. Screw gauge can measure the diameter of thin wires
or similar objects with accuracy upto
(A) 1 cm
(B) 0.1 cm
(C) 0.01 cm
(D) 0.001 cm
117. Pitch of screw of a screw gauge is the
(A)
distance moved by thimble on main scale
number of rotation off thimble
(B)
pitch
number of circular scale divisions
(C)
number of rotation of thimble
number of circular scale divissions
(D) None of the above
118. Least count of screw gauge is defined as
(A)
(B)
(C)
distance moved by thimble on main scale
number of rotation of thimble
pitch of the screw
number of circular scale
divisions & head
d scale
number of rotation of thimble
number of circular scale divissions
(D) None of the above
119. Screw gauge can be used to determine
(A) Thickness of glass piece
(B) Diameter of a wire
(C)To measure thickness of glass piece and to measure diameter of wire
(D) None of the above
120. Screw gauge contains following scales
(A) main scale, vernier scale
(B) main scale, ordinary scale
(C) main scale, head scale
(D) only main scale
121. Screw gauge is said to have negative error
(A)when head scale zeroth division coincides with
base line of main scale
(B)when head scale zeroth division is above with
base line of main scale
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2.51
Chapter 2: Measurements and General Physics
(C)when head scale zeroth division is below with
base line of main scale
(D) None of the above
122. For negative error correction is
(A) positive
(B) negative
(C) no correction
(D) None of the above
123. Screw gauge is said to have positive error
(A)when head scale zeroth division coincides with
base line of main scale
(B)when head scale zeroth division is above with
base line of main scale
(C)when head scale zeroth division is below with
base line of main scale
(D) None of the above
124. For positive error, the correction is
(A) positive
(B) negative
(C) nil
(D) None of the above
125. The diameter D of a wire is measured using screw
gauge of zero error. Then
⎛ main ⎞ ⎛ circular ⎞
⎛ leaast ⎞
(A) D = ⎜ scale ⎟ + ⎜ scale ⎟ × ⎜
⎟ ⎝ count ⎟⎠
⎟ ⎜
⎜
⎝ reading ⎠ ⎝ reading ⎠
(B)
⎛ circular ⎞ ⎛ main ⎞
⎛ leaast ⎞
D = ⎜ scale ⎟ + ⎜ scale ⎟ × ⎜
⎟ ⎝ count ⎟⎠
⎟ ⎜
⎜
⎝ reading ⎠ ⎝ reading ⎠
⎛ main ⎞ ⎛ vernier ⎞
⎛ leasst ⎞
(C) D = ⎜ scale ⎟ + ⎜ scale ⎟ × ⎜
⎟ ⎝ count ⎟⎠
⎟ ⎜
⎜
⎝ reading ⎠ ⎝ reading ⎠
(D) None of the above
126. The pitch of a screw gauge is 0.5 mm. Its head scale
contains 50 divisions. The least count of the screw
gauge is
(A) 0.01 mm
(B) 0.001 mm
(C) 0.02 mm
(D) 0.002 mm
127. Which of the following is correct?
(A)When the zero of the circular scale advances
beyond the reference line, then the zero correction is negative
(B)When the zero of the circular scale has advanced
beyond the reference line, then the zero correction is positive and when the zero of the circular
scale is left behind the reference line, then the
zero correction is negative
(C)When the zero of the circular scale is left behind
the reference line, then the zero correction is
positive
(D) Both (A) and (C)
02_Measurements, General Physics_Part 2.indd 51
128. The vernier of a circular scale has been divided into
30 divisions which coincide with 29 divisions of the
1°
main scale, having each division of
. The least
2
count of the device is
(A) 10′
(C) 0.1′
(B) 30′
(D) 1′
129. The velocity v of a particle at time t is given by
a
bt
. The dimensions of a , b , c are
v= + 2
t t +c
respectively
(B) L , L , T 2
(A) LT −2 , L , T
(C) L , LT , T −2
(D) L , L , LT 2
ky
,
y + a2
where U represents the potential energy, y represents the displacement and a represents the maximum displacement i.e., amplitude ?
130. What is the unit of k in the relation U =
(A) ms −1
(C) Jm
2
(B) ms
(D) Js −1
131. For a body moving along x-axis, the distance travelled by body from a reference point is given as function of time t as x = at 2 + b , where a and b are
constants, then the dimension of ab is same as
(A) speed
(B) distance travelled
(C) acceleration
(D) None of these
132. If a quantity x is defined by the equation x = 3CB2
where C is capacitance in farad and B represents
magnetic field in tesla. The dimensions of x are
(A) ML−2
(B) ML−2T −2 A
(C) ML−2T −2 A 2
(D) L−1 A −1
133. There are two different quantities A and B having
different dimensions. Then which of the following
operation is dimensionally correct?
(B) A − B
(A) A + B
A
(C)
(D) e A B
B
1 2
at where S is the distance
4
travelled, u is the initial velocity, a is the acceleration and t is the time is
(A) dimensionally correct only
(B) dimensionally incorrect only
(C) dimensionally and numerically correct
(D) dimensionally and numerically wrong
134. The formula S = ut −
135. The dimensional formula of a physical quantity x
is [ M −1L3T −2 ] . The error in measuring the quantities M , L and T are 2%, 3% and 4%. The maximum
11/28/2019 6:50:33 PM
2.52 JEE Advanced Physics: Mechanics – I
percentage of error that occurs in measuring, the
quantity x is
(A) 9
(B) 10
(C) 14
(D) 19
136. A particle of mass m is executing oscillations about
the origin on the x-axis. Its potential energy is
3
U ( x ) = K x where K is a positive constant. If the
amplitude of oscillation is a then its time period T
is
1
(B) Independent of a
(A) Proportional to
a
(C) Proportional to
a
(D) Proportional to a 3 2
137. In a particular system, the unit of length, mass
and time are chosen to be 10 cm , 10 g and 0.1 s
­respectively. The unit of force in this system will be
equivalent to
1
N
10
(C) 10 N
(B) 1 N
(A)
(D) 100 N
A−x
,
Bt
where x is in metre and t is in seconds. The dimensions of B is
138. Force acting on a particle is given by F =
(A) MLT −2
(B)
M −1T −3
(C) M −1T
(D) MT −1
139. The viscosity η of a gas depends on the long-range
attractive part of the intermolecular force, which
varies with molecular separation r according to
F = μr − n where n is a number and μ is a constant.
If η is a function of mass m of the molecules, their
mean speed v and the constant μ , then which of following is correct
(A) η ∝ m
n+1 n+ 3 n − 2
v
μ
n −n
−2
(C) η ∝ m v μ
(B)
n + 1 n + 3 −2
n −1
η ∝ m n −1 v n −1 μ
(D) η ∝ mvμ
−n
140. A science student takes 100 observations in an experiment. Second time he takes 500 observations in the
same experiment, by doing so the possible error
becomes
1
times
(B) 5 times
(A)
5
(C) unchanged
(D) None of these
h
( h = Planck’s constant and e =
e
electronic charge) are same as that of
(A) magnetic flux
(B) electric flux
(C) electric field
(D) magnetic field
141. The dimensions of
02_Measurements, General Physics_Part 2.indd 52
142.
E2
has the dimensions ( E = electric field, μ0 =
μ0
­permeability of free space)
[ M 2L3T −2 A2 ]
(C) [ ML3T −2 ]
(A)
[ MLT −4 ]
(D) [ M −1L2TA −2 ]
(B)
143. The dimensions of σ b 4 ( σ = Stefan’s constant and
b = Wein’s constant) are
[ M 0 L0T 0 ]
(C) [ ML−2T ]
(A)
[ ML4T −3 ]
(D) [ ML6T −3 ]
(B)
144. A quantity X is measured and expressed as X = Nu ,
where N numerical value and u is unit. Which
of the following is independent of system of unit
chosen?
(B) N × u
(A) N
N
(C)
(D) None of these
u
145. Which of the following combination of three quantities P , Q , R having different dimension cannot be
meaningful?
(B) PQ − 1
(A) PQ − R
(P −Q)
PR − Q 2
R
QR
146. A and B are two physical quantities having different dimensions. Then which of the following operation is dimensionally correct?
A
(B) log
(A) A + B
B
A
(C)
(D) e A B
B
(C)
(
(D)
)
147. For 10 at+ 3 , the dimension of a is
(A) M 0 L0T 0
(B)
M 0 L0T 1
(C) M 0 L0T −1
(D) None of these
148. Choose the wrong statement
ωL
(A)The dimensions of
are same as that of strain
R
1
(B)The dimensions of
are same as that of
LC
angular velocity
(C)The dimension of LCR are same as that of time
(D) None of these
149. After rounding off the number 4621 to 2 signification
digits the value becomes
(A) 4600
(B) 4500
(C) 4700
(D) 4720
150. Which of the following relation cannot be derived
using dimensional analysis (Neglect value of
constant)
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Chapter 2: Measurements and General Physics
(A) v =
F
μ
(B)
(C) F = 6πηrvd
T = 2π
(D) F =
l
g
Gm1m2
r2
151. Equation of plane progressive transverse wave in a
dissipative medium has general form
y = Ae −α x sin β ( t − Bx )
where α , A , B , C are constant, x and y are displacement and t is time. Dimensions of α , β and B
respectively are
(A)
(B)
(C)
(D)
M 0 L−1T 0 , M 0 L−1T −1 , M 0 LT −1
M 0 L1T 0 , M 0 L0T −1 , M 0 L−1T 1
M 0 L−1T 1 , M 0 L−1T , M 0 L0T −1
M 0 L−1T 0 , M 0 L0T −1 , M 0 L−1T 1
152. In a certain system of units, 1 unit of time is 5 sec ,
1 unit of mass is 20 kg and unit of length is 10 m .
In this system, one unit of power will correspond to
1
(A) 16 watt
(B)
watt
16
(D) None of these
(C) 25 watt
153. In book, the answer for a particular question is
2kl ⎤
ma ⎡
⎢ 1+
⎥ , where m represents
k ⎣
ma ⎦
mass, a represents acceleration, l represents length.
Then the unit of b should be
expressed as b =
(A) ms −1
(C) meter
(B) ms −2
(D) /sec
154. A student performs an experiment to determine the
Young’s modulus of a wire exactly 2 cm long, by
Searle’s method. In a particular reading, the student
measures the extension in the length of the wire to be
0.8 mm with an uncertainty of ±0.05 mm at a load
of exactly 1 kg . The student also measures the diameter of the wire to be 0.4 mm with an uncertainty
of ±0.01 mm . Take g = 9.8 ms −2 (exact). The Young’s
modulus obtained from the reading is
(A) ( 2 ± 0.3 ) × 1011 Nm −2
(B)
( 2 ± 0.2 ) × 1011 Nm −2
(C) ( 2 ± 0.1 ) × 1011 Nm −2
(D) ( 2 ± 0.05 ) × 1011 Nm −2
155. The time dependence of physical quantity P is given
2
by P = P0 e −α t + βt + γ , where α , β , γ are constants
and their dimensions are given by (where t is time)
02_Measurements, General Physics_Part 2.indd 53
2.53
(A) M 0 L0T −2 , M 0 L0T −1 , M 0 L0T 0
(B) M 0 L−1 , T −2 , M 0 L0T −1 , M 0 L0T
(C) M 0 L0T −1 , MLT −2 , M 0 L0T −1
(D) M , L , T , MLT 0 , M 0 L0T 0
156. The time period of a body under S.H.M. is represented by T = Pα Dβ Sγ , where P is pressure, D is
density and S is surface tension, then values of α ,
β and γ are
F
(Given that surface tension S = )
l
3 1
1
(A) − , , 1
(B) 1, 2,
2 2
3
(C) –1, –2, 3
(D)
1 −3 −1
,
,
2 2
2
157. The relative error in calculating the value of g from
l
the relation T = 2π
, if the relative errors in calcug
lating T and l are ±x and ±y respectively is
(B) 2x − y
(A) x + y
(C) 2x + y
(D) x − 2 y
e2
, where e , ε 0 , h and c
4πε 0 hc
are electronic charge, electric permittivity, Planck’s
constant and velocity of light in vacuum respectively is
158. The dimensions of
[ M 0 L0T 0 ]
(C) [ M 0 L1T 0 ]
(A)
[ M1L0T 0 ]
(D) [ M 0 L0T 1 ]
(B)
159. If E be the energy, G is the gravitational constant, I
be the impulse and M be the mass, then dimensions
GIM 2
of
are same as that of
E2
(A) time
(B) mass
(C) length
(D) force
160. The equivalent focal length of a combination of two
thin lenses (having focal length f1 and f 2 ) can be
1 1
1
given by
. If errors in measurement of f1
= +
f
f1 f 2
and f2 are ±0.5 cm and ±0.3 cm and given that f1 = f 2 .
Then errors in equivalent focal length is
(B) ±0.8
(A) ±0.1
(C) ±0.2
(D) ±0.4
161. If y represents distance and x represents time,
d2 y
dimensions of
are
dx 2
(B) L2T 2
(A) LT −1
2 −1
(C) L T
(D) LT −2
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2.54 JEE Advanced Physics: Mechanics – I
multiple correct choice type questions
This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE is/are correct.
1.
A student when discussing the properties of a medium
(except vacuum) writes
⎛ Velocity of light ⎞ ⎛ Velocity of light ⎞
=
⎜
⎝ in vacuum ⎟⎠ ⎜⎝ in medium ⎟⎠
This formula is
(A) dimensionally correct
(B) dimensionally incorrect
(C) numerically incorrect
(D) dimensionally and numerically correct
2.
The position of a particle moving along the y-axis is
given as y = At 2 − Bt 3 where y is measured in metre
and t in second. Then
(A) [ A ] = LT −2
(B) [ B ] = LT −3
(C)
[ A3 ]
=L
[ B2 ]
(D)
[ B3 ]
=L
[ A2 ]
3.
Dimensional analysis cannot be used to derive
formulae
(A) containing trigonometrical functions
(B) containing exponential functions
(C) containing logarithmic functions
(D) containing dimensionless quantities
4.
Which of the following is (are) dimensionless?
(A) Refractive index
(B) Poisson’s ratio
(C) Universal gravitational constant
(D) Specific gravity
5.
6.
7.
The
equation of the stationary wave is
⎛ 2π ct ⎞
⎛ 2π x ⎞ . Which of the following
y = 2 A sin ⎜
cos ⎜
⎝ λ ⎟⎠
⎝ λ ⎟⎠
statement(s) is/are correct?
(A) The unit of ct is same as that of λ
(B) The unit of x is same as that of λ
(C) The unit of 2π c λ is same as that of 2π x λt
(D) The unit of c λ is same as that of x λ
Which of the following can be expressed as
dynecm −2 ?
(A) Pressure
(B) Longitudinal stress
(C) Young’s modulus of elasticity
(D) Viscosity
The pair(s) having same dimensions is/are
(A) Torque and work
(B) Angular momentum and work
02_Measurements, General Physics_Part 2.indd 54
(C) Energy and Young’s modulus
(D) Light year and wavelength
8.
The pairs of physical quantities that possess same
dimensions is/are
(A) Renold’s number and coefficient of friction
(B) Curie and frequency of light wave
(C) Latent heat and gravitational potential
(D) Planck constant and torque
9.
x
z
If dimensions of length are expressed as G c h ,
where G , c and h are the universal gravitational
constant, speed of light and Planck constant respectively, then
y
(A) x =
1
1
, y=
2
2
3
1
(C) y = − , z =
2
2
x=
1
1
,z=
2
2
(D) y =
1
3
, z=
2
2
(B)
10. Choose the correct statement(s)
(A)A dimensionally correct equation may be correct
(B)A dimensionally correct equation may be incorrect
(C)
A dimensionally incorrect equation may be
correct
(D)
A dimensionally incorrect equation must be
incorrect
11. Out of the following unit(s), the one(s) measuring
energy is/are
2
−1
(A) kWhr
(B) ( volt ) ( sec )( ohm )
(C)
( pascal ) ( foot )2
(D) ( weber ) ( ampere )
E
1
and
, y=
B
μ0 ε 0
l
z=
. Here l is the length of a wire, C is a capaciCR
tance and R is a resistance. All other symbols have
standard meanings
(A) x , y have the same dimensions
(B) x , z have the same dimensions
(C) y , z have the same dimensions
(D)None of the three pairs have the same dimensions
12. Consider three quantities; x =
13. Dimensions of light year are the same as those of
(A) leap year
(B) wavelength
(C) radius of gyration
(D) propagation constant
14. Which of the following is/are not a unit of time?
(A) parsec
(B) light year
(C) micron
(D) second
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Chapter 2: Measurements and General Physics
15. Choose the pairs of physical quantities, which have
identical dimensions.
(A) Impulse and linear momentum
(B) Planck constant and angular momentum
(C) Moment of inertia and moment of force
(D) Young’s modulus and pressure
16. The dimensions of energy per unit volume are the
same as those of
(A) work
(B) stress
(C) pressure
(D) modulus of elasticity
17.
⎛ B2 ⎞
⎜⎝ μ ⎟⎠ has the same dimensional formula as that of
0
( B is the magnetic field induction and μ0 is absolute
permeability of free space)
(A) energy density
(B) magnetic energy per unit volume
(C) magnetic intensity
(D) intensity of magnetisation
18. Which of the following is/are correct?
(A)The term science is derived from a Latin verb
which means ‘to know’
(B)The term physics in derived from a Greek word
which means ‘nature’
(C) L is the symbol of unit of a physical quantity
(D)Physics is an exact science because it is based on
the experimental measurements
19. In international system of units, there are seven base
quantities whose units are defined. Which physical
quantities do not have a prefix with their units?
(A) Amount of substance
(B) Thermodynamic temperature
(C) Luminous intensity
(D) Mass
20. Which of the following is/are the units of mass?
(A) kgf
(B) metric ton
(C) quintal
(D) amu
21. Which of the following can be expressed as Nm −2 ?
(A) Energy density of electric field
(B) Bulk modulus of elasticity
(C) Pressure of 1 mm mercury column
(D) Compressional stress
22. Which of the following physical quantities have
dimensions zero in mass, 2 in length and −2 in time?
(A) Latent heat
(B) Potential energy per unit mass
(C) Gravitational potential
(D) Spring constant
02_Measurements, General Physics_Part 2.indd 55
2.55
23. Which of the following is/are true regarding significant figures?
(A) All non-zero digits are significant
(B)The zeros appearing in the middle of a number
are significant, while those at the end of a number
without a decimal point are ambiguous
(C)The powers of 10 are counted while counting the
number of significant figures
(D)
Greater the number of significant figures in a
measurements smaller is the percentage error
24. Which of the following cannot be considered to be the
most accurate?
(B) 20 × 101 m
(A) 200 m
(C) 2 × 10 2 m
(D) Data insufficient
25. The focal length ( f ) of a curved mirror is related to
the object distance u and the image distance v in
1 1 1
accordance with the mathematical relation, = + .
f u v
The maximum relative error in calculating the focal
length f of the mirror is
(A)
Δf
f
2
=
Δu
u
2
+
Δv
v2
(B)
Δf
1
1
=
+
f
Δu u Δv v
(C)
Δf
uv ⎛ Δv Δv ⎞
+
=
⎜
⎟
f
u + v ⎝ v2 v2 ⎠
(D)
Δf Δu Δv Δu
Δv
=
+
+
+
f
u
v u+v u+v
t
26. In the equation, y = a cos ⎛ − qx ⎞ , where t represents
⎜⎝ p
⎟⎠
time in second and x represents distance in metre.
Which of the following statement(s) is/are false?
(A) The unit of x is same as that of q
(B) The unit of x is same as that of p
(C) The unit of t is same as that of q
(D) The unit of t is same as that of p
⎡ 2π (
⎤
27. In the relation, y = A sin ⎢
ct − x ) ⎥ , where y and
⎣ λ
⎦
x are measured in metre. Which of the following
statement(s) is/are false?
(A) The unit of λ is same as that of x and A
(B)The unit of λ is same as that of x but not of A
(C) The unit of c is same as that of x
2π
(D) The unit of ( ct − x ) is same as that of
λ
11/28/2019 6:51:13 PM
2.56 JEE Advanced Physics: Mechanics – I
28. Identify the pair(s) having identical dimensions.
(A) Linear momentum and moment of force
(B) Planck constant and angular momentum
(C) Pressure and modulus of elasticity
(D) Work and torque
29. Which of the following is/are true?
(A) The order of magnitude of 501 is 3
(B) The order of magnitude of 499 is 2
(C) The order of 230 is nearly 1090
(D)
The unit of reduction factor of a tangent
­galvanometer is ampere
where, all the primed symbols belong to one system of
units and the unprimed symbols belong to the other
system of units. Here α and β are dimensionless constants. Which of the following is/are correct?
(A)
The standard of length in each of these two
α3
­systems are related to each other as l′ = 3 l
β
(B)The standard of mass in each of these two ­systems
⎛ 1 ⎞
are related to each other as m′ = ⎜ 2 2 ⎟ m
⎝α β ⎠
(C)The standard of time in each of these two systems
⎛ α ⎞
are related to each other as t ′ = ⎜ 2 ⎟ t
⎝β ⎠
30. The velocity, acceleration and force in two systems of
units are related to each other as under
(i) v ’ =
α2
v
β
(ii)
(D)The standard of momentum in each of these two
⎛ 1 ⎞
systems are related to each other as p′ = ⎜ 3 ⎟ p
⎝β ⎠
a′ = ( αβ ) a
1 ⎞
(iii) F ′ = ⎛⎜
F
⎝ αβ ⎟⎠
reasoning based questions
This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as
Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.
Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1.
Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.
Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.
1.
Statement-1: The light year and wavelength have
dimensions of length.
Statement-2: Both light year and wavelength represent distances.
2.
Statement-1: The fundamental units selected in a particular system are the velocity of light, 3 × 108 ms −1 ,
acceleration due to gravity is 10 ms −2 and the mass
of proton is 1.67 × 10 −27 kg .
Statement-2: The value of time in such a system is
3 × 107 s .
3.
Statement-1: Surface energy stands for energy and not
for energy per unit area.
Statement-2: Dimensional formula of surface energy is
4.
⎡ MT −2 ⎤ .
⎣
⎦
Statement-1: The density of earth is given by
ρ=
3g
, where Re stands for the radius of earth.
4π Re G
Statement-2: g = GM .
Re2
02_Measurements, General Physics_Part 2.indd 56
5. Statement-1: The value of 1 micron is equal to 10 −6 m.
Statement-2: Micron is the unit for measuring microscopic distance.
6.
Statement-1: The dimensional formula for electro-
static potential is ⎡ ML2T −3 A −1 ⎤ .
⎣
⎦
Statement-2: Electrostatic potential is directly proportional to work done.
7.
Statement-1: Distance travelled in a particular second
has the dimensions same as that of distance.
Statement-2: Velocity is the displacement covered per
unit time.
8.
Statement-1: When percentage errors in the measurement of mass and velocity are 1% and 2% respectively,
the percentage error in the measure of kinetic energy
( E ) is 5%.
Statement-2: The relative error in Kinetic Energy or
ΔE Δm 2 Δv
1
=
+
.
E = mv 2 is
E
m
v
2
11/28/2019 6:51:16 PM
Chapter 2: Measurements and General Physics
2.57
1
is numerically
μ oε o
equal to velocity of light and has dimensional formula
same as that of velocity.
Statement-2: μ o is permeability of free space and ε o is
the permittivity of free space.
18. Statement-1: Avogadro number is a dimensionless
constant.
Statement-2: It is number of atoms in 1 gram mole.
10. Statement-1: Specific gravity is measured by
hygrometer.
Statement-2: Thermometer is an instrument which is
used to measure the temperature of a body.
Statement-2: According to the principle homogeneity
of dimensional formula, only that formula is correct,
where the dimensions of left hand side is equal to the
dimensional formula of right hand side.
11. Statement-1: The error in the measurement of radius
of the sphere is 0.3%. The permissible error in its surface area is 0.6%.
Statement-2: The permissible relative error is calcuΔA
Δr
lated by the formula
=4 .
A
r
12. Statement-1: Pressure can be subtracted from the pressure gradient.
Statement-2: Only like quantities can be added or subtracted from each other.
20. Statement-1: AU is a much bigger unit than Å .
9.
Statement-1: The quantity
13. Statement-1: Method of dimensions cannot be used
for deriving formula containing trigonometrical ratios.
Statement-2: This is because trigonometrical ratios
have no dimensions.
14. Statement-1: A fundamental quantity is the one that
does not depend upon other quantities.
Statement-2: Length, mass and time are derived
quantities.
15. Statement-1: In any mathematical relation between
physical quantities the dimensions on both the sides
must be the same.
Statement-2: The dimensions of a physical quantity
are the powers raised on fundamental units to obtain
the derived units of that physical quantity.
16. Statement-1: The unit used for measuring nuclear
cross section is barn.
Statement-2: 1 barn = 10 −4 m 2
1 T
where
2l m
­symbols have usual meanings, m represents linear
mass density.
Mass
Statement-2: Linear mass density is m =
Volume
17. Statement-1: In the relation,
02_Measurements, General Physics_Part 2.indd 57
f =
19. Statement-1: The time period of a simple pendulum is
l
given by the formula, T = 2π
.
g
Statement-2: 1 AU = 1.5 × 1011 m and 1 Å = 10 −10 m
21. Statement-1: The mass of largest stone that can be
moved by flowing river depends on velocity v of
river, density ρ and acceleration due to gravity g .
Statement-2: It can be shown that m is proportional to
the sixth power of v .
22. Statement-1: Pressure has the dimensions same as that
of energy density.
Energy
Statement-2: Energy density =
=Pressure
Volume
23. Statement-1: The graph between P and Q is a
P
= constant.
straight line, when
Q
Statement-2: The straight line graph means the slope
of the graph is constant.
Δx
Δa
Δb
Δc
a b m
=
+m
+n
,
, then
n
x
a
b
c
c
Δa Δb
Δc
where
,
and
are the fractional errors in the
a
b
c
values of a , b and c, respectively.
Statement-2: The above relation is valid only when
Δa a , Δb b and Δc c .
24. Statement-1: If x =
Δx
Δa
Δb
Δc
a b m
=
+m
−n
,
, then
n
x
a
b
c
c
where Δa , Δb and Δc are the increments in the values of a , b and c, respectively.
Statement-2: The above relation is valid only when
Δa a , Δb b and Δc c .
25. Statement-1: If x =
11/28/2019 6:51:22 PM
2.58 JEE Advanced Physics: Mechanics – I
linked comprehension type questions
This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph
followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of
competitiveness there may be a few questions that may have more than one correct options)
Comprehension 1
The following set of quantities have been observed to possess the same dimensional formulae.
A :Velocity, speed, distance covered in nth second.
B :
Acceleration,
retardation,
gravitational
_______________.
C :
Work, energy, heat, internal energy, moment of
_______________.
D :Force, thrust, weight, energy _______________.
E :
Surface tension, surface _______________, spring
constant.
F :
Pressure, stress, _______________ density, Young’s
modulus, Bulk modulus.
Based on the above information answer the following
questions.
1.
In B, the blank space is best filled with
(A) field
(B) potential
(C) intensity
(D) force
2.
In C, the blank space is best filled with
(A) torque
(B) force
(C) momentum
(D) velocity
3.
In D, the blank space is best filled with
(A) density
(B) rate
(C) gradient
(D) None of these
4.
In E, the blank space is best filled with
(A) density
(B) energy
(C) area
(D) None of these
5.
In F, the blank space is best filled with
(A) energy
(B) number
(C) mass
(D) force
Comprehension 2
a
The vander Waals equation is ⎛⎜ p + 2 ⎞⎟ ( V − b ) = RT ,
⎝
V ⎠
where p is pressure, V is molar volume and T is the temperature of the given sample of gas. R is called molar gas
constant, a and b are called vander Wall constants.
Based on the above information answer the following
questions.
6.
The dimensional formula for b is same as that for
(B) V
(A) p
(D) RT
(C) pV 2
02_Measurements, General Physics_Part 2.indd 58
7.
The dimensional formula for a is same as that for
(B) p
(A) V 2
(C) pV 2
(D) RT
8.
Which of the following does not possess the same
dimensional formula as that for RT ?
(B) pb
(A) pV
(C)
9.
a
V2
(D)
ab
V2
ab
is
RT
(B) M 0 L3T 0
(D) M 0 L6T 0
The dimensional formula for
(A) ML5T −2
(C) ML−1T −2
10. The dimensional formula of RT is same as that of
(A) energy
(B) force
(C) specific heat
(D) latent heat
Comprehension 3
Dimensional analysis is a good method to find the dependency of a physical quantity on other physical quantities.
Assume that the velocity of light c , universal gravitational
constant G and the Plank’s constant h be chosen as the
fundamental units. Based on the above facts, answer the
following questions.
11. In this new system, mass will have a dimensional formula given by
1
−
1
1
(A) c 2 G 2 h 2
−
5
1
(B)
−
3
1
1
c 2G 2 h 2
1
(C) c 2 G 2 h 2
(D) None of these
12. In this new system, length will have a dimensional formula given by
1
−
1
1
(A) c 2 G 2 h 2
−
5
1
(B)
−
3
1
1
c 2G 2 h 2
1
(C) c 2 G 2 h 2
(D) None of these
13. In this new system, time will be expressed as
1
−
1
1
(A) c 2 G 2 h 2
−
5
1
(B)
−
3
1
1
c 2G 2 h 2
1
(C) c 2 G 2 h 2
(D) None of these
11/28/2019 6:51:28 PM
2.59
Chapter 2: Measurements and General Physics
Comprehension 4
Measurements always have uncertainties. If you measure
the thickness of the cover of the book “MECHANICS-I by
RAHUL SARDANA” using an ordinary ruler, then your
measurement may be reliable only to the nearest millimeter
and your result may be 2 mm (say). It would be absolutely
wrong to state this result as 2.00 mm till we are not aware
of the limitations of the measuring device. Based on the
above facts, answer the following questions.
14. The percentage error in measuring a distance of about
50 cm with a meter stick having callibrations upto
1 mm is
(A) 0.1%
(B) 0.2%
(C) 0.3%
(D) 0.4%
15. The percentage error in measuring a mass of about 1 g
with a balance having a least count 0.1 mg is
(A) 0.04%
(B) 0.03%
(C) 0.02%
(D) 0.01%
16. The percentage error in measuring a time interval of 4
minute with a stop watch having least count of 0.2 s is
(A) 5%
(B) 0.06%
(C) 0.08%
(D) 0.10%
Comprehension 5
The following set of quantities have been observed to possess the same dimensional formulae.
A :_______________, Planck’s constant and energy per
unit _______________.
B :
Frequency,
angular
frequency,
angular
_______________ and _______________ gradient.
C :_______________ field strength and _______________
potential gradient.
D :_______________ and latent heat.
E :_______________ and strain.
Based on the above information answer the following
questions.
(C) energy at first blank, speed at second blank
(D) energy at both blanks
19. In C, the blank space is best filled by
(A) electric at both
(B) electric at first, gravitational at second
(C) gravitational at first, electric at second
(D) gravitational at both
20. In D, the blank space is best filled by
(A) gravitational potential
(B) kinetic energy per unit mass
(C) gravitational potential energy per unit mass
(D) None of these
21. In E, the blank space is best filled by
(A) angle
(B) velocity of light
(C) fine structure constant (D) Plank’s constant
Comprehension 6
A particle is moving along a straight line and variable force
F is acting on this F at any time t is given by
B
Ct
+
t D + t2
where A , B , C and D are constant. Based on the above
facts, answer the following questions.
F = A+
22. The dimensional formula of C is
[ MLT –2 ]
(C) [ MLT –3 ]
(A)
(B)
[ MLT –1 ]
(D) [ ML ]
23. The dimensions of B is equal to dimensions of
(A) Force
(B) Power
(C) Linear momentum
(D) Energy
24. Dimensions of ⎛⎜ 1 ⎞⎟ is equal to dimension of
⎝ D⎠
(A) frequency
(B) time
(C) length
(D) force
Comprehension 7
17. In A, the blank space is best filled with
(A) momentum in both blanks
(B)energy in first blank and angular momentum in
the second blank
(C)angular momentum in first blank and frequency
in the second blank
(D)frequency in the first blank and angular momentum in the second blank
Angular velocity of ceiling fan depends upon (i) applied
voltage (V ), (ii) resistance (R) of the coil of fan, (iii) mass
(m) of fan and (iv) length (l) of the fan-blades. Given that
the dimensional formula for voltage V is ML2T −3 A −1 and
V
the resistance is defined as R = , where i is the current
i
flowing. Based on the above facts, answer the following
questions.
18. In B, the missing blank(s) are best filled as
(A) speed at both blanks
(B) speed at first blank, energy at second blank
25. Angular velocity of fan is proportional to
(A) R −1 3
(B) R −2 3
02_Measurements, General Physics_Part 2.indd 59
(C) R −3 2
(D) None of these
11/28/2019 6:51:31 PM
2.60 JEE Advanced Physics: Mechanics – I
26. By what factor angular velocity of fan will increase if
applied voltage is increase 8 times
(A) 8
(B) 4
(C) 2
(D) 3
27. Length of blade of fan is increased from 10 inch to
14.4 inch keeping mass and coil-resistance same. How
many times applied ­voltage needs to be increased as to
keep angular velocity constant
(A) 1.44 times
(B) 1.25 times
(C) 2.07 times
(D) 1.2 times
The potential energy of a particle varies with distance x
A x
, where A and B are
x+B
constants. Based on the above facts, answer the following
questions.
from a fixed origin as U =
28. The dimension of AB is
(A) ML5 2T −2
(B)
(C) ML1 2T −2
(D) ML7 2T −2
(A) ML T −2
32
−1 2
(C) ML
T
(A) M 2 L2T −2
2 4
(C) M L T
A
is
B2
−2
ML3 2T −2
(B)
ML T −2
12
A2
is
B
(B)
M 2 L2T −4
(D) M 2 L−4T −4
If energy ( E ) , velocity ( V ) and force ( F ) be taken as fundamental quantities. Based on the above facts, answer the
following questions.
31. Dimensional formula for surface tension is
(A) E1V −2
(B)
(C) E −1VF 2
(D) E1V 2
E −1F 2
32. Dimensional formula for angular momentum is
(A) E −2V 1F1
(B) E −2V 1F −1
2 −1 1
(C) E V F
(D) E2V −1F −1
33. Dimensional formula for time is
02_Measurements, General Physics_Part 2.indd 60
(A) −
5
6
1
2
35. The value of b is
(B) EV 2 F −1
(D) EVF 2
(B)
1
2
(D) 1
(A) −
5
6
(B)
(C) −
1
2
(D) 1
1
2
36. The value of c is
(A) −
5
6
1
3
(B)
1
2
(D) 1
Comprehension 11
If Force ( F ) , acceleration ( A ) and time ( T ) is considered
to be fundamental quantities. Based on the above facts,
answer the following questions.
37. The dimensions of torque are
(A) FA −1T 2
(C) FAT −2
Comprehension 9
(A) EV −1F −1
(C) EV –1F
34. The value of a is
(C)
(D) ML−1 2T 2
−2
30. The dimension of
A gas bubble, from an explosion under water, oscillates
with a period T proportional to p a db Ec , where p is the
static pressure, d is the density of water and E is the total
energy of explosion. Based on the above facts, answer the
following questions.
(C) −
Comprehension 8
29. The dimension of
Comprehension 10
(B) FA 2T −1
(D) FAT 2
38. The dimensions of velocity are
(A) AT
(B) AT 2
2
(C) A T
(D) FA 2T
39. The dimensions of momentum are
(A) FT 2
(B) F 2T
(C) FT
(D) FT −1
Comprehension 12
The distance travelled by a particle is given by
At ⎞
⎛
y = D ln ⎜ 1 + 2 ⎟ + B sin ( Cx 2 )
⎝
λ ⎠
here x and y are length, l is wavelength and t is time. Based
on the above facts, answer the following questions.
11/28/2019 6:51:41 PM
Chapter 2: Measurements and General Physics
2.61
Comprehension 14
40. The dimensions of A and C are
(A) L2T −1 , L2
(B) L2T −1 , L−2
(C) L−2T 1 , L−2
(D) L−2T −1 , L2
The equation of state of gas is
a ⎞
⎛
⎜⎝ P + 2 ⎟⎠ ( V − b ) = RT
V
41. Which pair has same dimensions
(B) B, C
(A) A , B
(D) B, D
(C) C , D
Comprehension 13
here, P is the pressure, V is the volume, T is the absolute
temperature and a , b and R are constants. Based on the
above facts, answer the following questions.
If pressure of a gas is given by
44. The dimensions of a are
P=
(A) ML5T −2
2
α
⎛ αh ⎞
sin ⎜
β
⎝ kBθ ⎟⎠
(B) ML5T 2
(C) MLT
(D) M 2 L5T −2
where h is the Plank’s constant
kB is the Boltzmann constant
θ is the absolute temperature
45. The dimensions of b are
(A) MLT
Based on the above facts, answer the following questions.
(B)
42. The dimensions of β are
[ M1L0T 0 ]
(C) [ M −1LT 0 ]
(A)
(B)
M 0 L3T 0
(C) ML3T 0
(D) ML3T 1
[ M 0 L0T 0 ]
(D) None of these
46. The dimensions of R are
43. Which of the following is not dimensionless
βP
(A) α t
(B)
α2
h
BI
(C)
(D) 2
kθ
L
( I : Moment of inertia, L : length and t : time)
(A) ML2T −2 K
(B)
ML2T −2 K −1
(C) MLT −2 K −1
(D) MLT −2 K
Matrix Match/Column Match Type Questions
Each question in this section contains statements given in two columns, which have to be matched. The statements in
COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of
bubbles will look like the following:
A
B
C
D
02_Measurements, General Physics_Part 2.indd 61
p
p
p
p
p
q
q
q
q
q
r
s
t
r
r
r
r
s
s
s
s
t
t
t
t
11/28/2019 6:51:48 PM
2.62 JEE Advanced Physics: Mechanics – I
1.
Match the following
Column-I
Column-II
R
L
(p) Time
(A)
(B) CR
5.
(C)
E
B
(r) Speed
(D)
1
μ0 ε 0
(s) Kinetic energy
Match the following
Column-I
Column-II
(A) Stress
(p) Pressure
(B) Strain
(q) Energy density
(C) Modulus of elasticity
(r) Angle
(D) Torque
(s) Energy
(s) Linear momentum
2mE
6.
Column-II
2
−2
(p) ML T K
(A) Specific heat
−3
Match the units given in COLUMN-I with the physical quantities in COLUMN-II
Column-I
Column-II
(A) Nm −2
(p) Force constant
(B) Nm −1
(q) Surface energy of a liquid
(C) Nm
(r) Stress
(D) kgs −2
(s) Bulk modulus
(t) Torque
Match the following
Column-I
Match COLUMN-I with COLUMN-II
Column-I
Column-II
(A) Dimensionless
quantity
(p) Angle
(B) Young’s
modulus
(q) Mechanical equivalent
of heat
(C) Jcal −1
(r) kgm −1s −2
(D) pascal
(s) Thermal conductivity
−1
(t) Fine structure constant (a)
−4
(B) Boltzmann’s constant
(q) MT K
(C) Wien’s constant
(r) LK
(D) Stefan’s constant
(s) L2T −2 K −1
7.
Match the units/dimensions in COLUMN-I with the
physical quantities/expressions in COLUMN-II
Column-I
Column-II
(A) Jkg −1
(p)
Match the units/dimensions in COLUMN-I with the
physical quantities in COLUMN-II
(B) M 0 L2T −2 K −1
(q) Mean square velocity
Column-I
Column-II
(C) M 0 L2T −2
(r) Latent heat
(A) ML2T −1
(p) Impulse
(D) JK −1
(s) Specific heat
(B) Js
(q) Planck’s constant
(C) MLT −1
(r) Angular momentum
(t) ML2T −3 K −4
4.
(D) Energy per unit
frequency
(t)
(t) Fine structure
constant
3.
Column-II
(q) Frequency
(t) Kinetic energy per unit mass
2.
Column-I
k BT
m
(t) Entropy
(Continued)
02_Measurements, General Physics_Part 2.indd 62
11/28/2019 6:51:51 PM
Chapter 2: Measurements and General Physics
8.
10. Match the following
Match the following
Column-I
Column-II
Column-I
Column-II
(A) Stress
(p) Pressure
(A) Young’s modulus
(p) L2T −2
(B) Strain
(q) Energy density
(B) Gravitational potential
(q) M −1L3T −2
(C) Modulus of elasticity
(r) Angle
(C) Latent heat
(r) ML−1T −2
(D) Torque
(s) Energy
(D) Gravitational constant
(s) ML2T −2
(t) Renold’s
number (nR)
9.
2.63
11. Match the following
(All symbols used have their standard meanings)
Column-I
Column-II
Some physical quantities are given in COLUMN-I and
some possible SI units in which these quantities may
be expressed are given in COLUMN-II. Match the
physical quantities in COLUMN-I with the units in
COLUMN-II.
(A) Coefficient of viscosity
(p) Dimensionless
(B) Strain
(q) Unitless
(C) Angle
(r) ML−1T −1
(D) Stress
(s) ML−1T −2
Column-I
Column-II
(A) GMe Ms
(p) (volt) (coulomb)
(metre)
G–universal
gravitational constant,
Me–mass of the earth,
Ms–mass of the sun.
(B)
3RT
(q) ( kilogram )
M
( metre )3
R–universal gas constant,
( second )−2
T–absolute temperature,
M–molar mass.
F2
(C) 2 2
qB
GMe
Re
G–universal gravitational
constant,
Me–mass of the earth,
Re–radius of the earth.
02_Measurements, General Physics_Part 2.indd 63
Column-I
( second )−2
e2
2 hε 0c
(p) Joule
(B)
R 2C 2
μ0 ε 0
(q) Dimensionless
(C) kBT
(s) ( farad )
( volt )2
( kg )−1
Column-II
(A)
(r) ( metre )2
F–force,
q–charge,
B–magnetic field.
(D)
12. In the column given below:
C stands for capacitance
R stands for resistance
k stands for Boltzmann constant
c
stands for speed of light
e
stands for electronic charge
H stands for Henry
(D)
e 4m
8ε 02 h 3 c
(r) m −1
(s) Unitless
13. Match the physical quantities given in COLUMN-I
with dimensional formula expressed in COLUMN-II
in tabular form
11/28/2019 6:51:55 PM
2.64 JEE Advanced Physics: Mechanics – I
17. Match the following
Column-I
Column-II
(A) Coefficient of viscosity
(p) MLT −1
Column-I
Column-II
(B) Angular momentum
(q) ML2T −2
(A) Coefficient of viscosity
(p) Dimensionless
(C) Torque
(r) LT −1
(B) Strain
(q) Unitless
(C) Angle
(r) ML−1T −1
(D) Stress
(s) ML−1T −2
(D)
1
μ0 ε 0
−1 −1
(s) ML T
14. Match the physical quantities given in COLUMN-I
with unit expressed in COLUMN-II in tabular form
Column-I
Column-II
(A) Pressure
(p) Watt
(B) Power
(q) Pascal
(C) Charge
(r) Hertz
(D) Frequency
(s) Coulomb
15. Match the following
Column-I
Column-II
(A) Young’s modulus
(p) L2T −2
−1 3
−2
(B) Gravitational potential
(q) M L T
(C) Latent heat
(r) ML−1T −2
(D) Gravitational constant
(s) ML2T −2
16. Match the following
18. The Stefan-Boltzman’s constant is not a fundamental
constant and one can write it in terms of fundamental
constant and written as σ = ahα C β Gγ kBδ , a is dimensionless constant, h is planck constant, C is speed of
light, G is universal gravitational constant and kB is
Boltzman constant.
Column-I
Column-II
(A) a
(p) -2
(B) b
(q) -3
(C) g
(r) 4
(D) d
(s) 0
19. Match COLUMN–I with COLUMN–II in regard
to dimensions of physical quantities mentioned in
COLUMN–I and dimensions of physical quantities in
COLUMN–II.
Column-I
Column-II
(A) Kinetic energy
(p) Pressure
(B) Energy density
(q) Torque
Column-I
Column-II
(C) Planck constant
(r) Angular momentum
(A) Electric potential
(p) ML2T −2 A −2
(D) Efficiency of carnot
cycle
(s) Longitudinal
strain
(B) Resistance
(q) ML2T −3 A −1
(C) Capacitance
(r) ML2T −3 A −2
(D) Inductance
(s) M −1L−2T 4 A 2
02_Measurements, General Physics_Part 2.indd 64
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Chapter 2: Measurements and General Physics
2.65
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data
given in the question(s).
1.
2.
EJ 2
, where E , M , J
M 5G 2
and G denote energy, mass, angular momentum and
gravitational constant is M a LbT c . Then find values of
a, b, c .
ex
has dimensional formula M 0 L0T 0 .
2 hε 0c
The quantity
Calculate x .
3.
20 N , 200 J and 5 ms −1 . Calculate the units of mass,
length and time in this new system.
The dimensional formula of
The rate of flow ( V ) of a liquid flowing through a
pipe of radius r and pressure gradient p is given by
Poiseuille’s equation
π rx y
p
8 ηz
Then calculate x , y , z .
5.
Consider a new system of units in which the unit of
mass is α kg , unit of length is β m and that of time
is γ s . The value of calorie in the new system is calculated to be 4.2α − x β − yγ z . Find the values of x , y , z .
6.
According to Kepler’s Laws of planetary motion,
the planets move around the sun in nearly circular
orbits. Assuming that the period of rotation depends
upon radius of the orbit ( r ) , mass of sun ( M ) and
Universal Gravitational Constant ( G ) as
V=
4.
A new system is formed such that it uses force, energy
and velocity as fundamental quantities with units
T a = 4π 2
rb
M cG d
Find a , b , c and d .
ARCHIVE: JEE MAIN
1.
[Online April 2019]
In SI units, the dimensions of
2
−1 −1
−1
3
(A) AT M L
(C) A TML
(B)
ε0
is
μ0
AT
−3
3
ML2
(D) A 2T 3 M −1L−2
2. [Online April 2019]
If Surface tension ( S ) , Moment of inertia ( I ) and
Planck’s constant ( h ) , were to be taken as the fundamental units, the dimensional formula for linear
momentum would be
(A)
1 3
S 2 I 2 h −1
3 1
S 2 I 2 h0
(B)
(C)
1 1
S 2 I 2 h −1
(D) S 2 I 2 h0
1 1
3. [Online April 2019]
In a simple pendulum experiment for determination
of acceleration due to gravity ( g ) , time taken for
20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes
out to be 30 s . The length of pendulum is measured
by using a meter scale of least count 1 mm and the
value obtained is 55.0 cm . The percentage error in the
determination of g is close to
02_Measurements, General Physics_Part 2.indd 65
(A) 6.8%
(B)
(C) 3.5%
(D) 0.7%
0.2%
4. [Online April 2019]
In the density measurement of a cube, the mass and
edge length are measured as ( 10.00 ± 0.10 ) kg and
( 0.10 ± 0.01 ) m , respectively. The error in the measurement of density is
(A) 2100 kgm −3
(B)
3100 kgm −3
(C) 6400 kgm −3
(D) 1100 kgm −3
5. [Online April 2019]
The area of a square is 5.29 cm 2 . The area of 7 such
squares taking into account the significant figures is
(A) 37.03 cm 2
(B)
37.0 cm 2
(C) 37.030 cm 2
(D) 37 cm 2
6. [Online April 2019]
In the formula X = 5YZ 2 , X and Z have dimensions
of capacitance and magnetic field, respectively. What
are the dimensions of Y in SI units?
[ M −2L−2T 6 A3 ]
(C) [ M −2 L0T −4 A −2 ]
(A)
[ M −1L−2T 4 A2 ]
(D) [ M −3 L−2T 8 A 4 ]
(B)
11/28/2019 6:52:13 PM
2.66 JEE Advanced Physics: Mechanics – I
7. [Online April 2019]
Which of the following combinations has the dimension of electrical resistance ( ε 0 is the permittivity of
vacuum and μ0 is the permeability of vacuum)?
(A)
ε0
μ0
(B)
ε0
μ0
(C)
μ0
ε0
(D)
μ0
ε0
8. [Online January 2019]
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and
100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale
lies 3 divisions below the mean line. The readings of
the main scale and the circular scale, for a thin sheet,
are 5.5 mm and 48 respectively, the thickness of this
sheet is
(B) 5.740 mm
(A) 5.725 mm
(C) 5.755 mm
(D) 5.950 mm
9. [Online January 2019]
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of
light) is proportional to
3
(A)
Gh
c5
(B)
c
Gh
(C)
Gh
c3
(D)
hc 5
G
10. [Online January 2019]
The density of a material is SI units is 128 kgm −3 . In
certain units in which the unit of length is 25 cm and
the unit of mass is 50 g , the numerical value of density of the material is
(A) 640
(B) 410
(C) 40
(D) 16
11. [Online January 2019]
The diameter and height of a cylinder are measured
by a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm,
respectively. What will be the value of its volume in
appropriate significant figures?
(A) 4264 ± 81 cm 3
(B)
4300 ± 80 cm 3
(C) 4260 ± 80 cm 3
(D) 4264.4 ± 81.0 cm 3
12. [Online January 2019]
The force of interaction between two atoms is given by
⎛ x2 ⎞
F = αβ exp ⎜ −
; where x is the distance, k is the
⎝ α kt ⎟⎠
Boltzmann constant and T is temperature and α and
β are two constants. The dimension of β is
02_Measurements, General Physics_Part 2.indd 66
(A) M 0 L2T −4
(B)
M 2 LT −4
(C) MLT −2
(D) M 2 L2T −2
13. [Online January 2019]
If speed ( V ) , acceleration ( A ) and force ( F ) are
considered as fundamental units, the dimension of
Young’s modulus will be
(A) V −2 A 2 F −2
(B) V −2 A 2 F 2
(C) V −4 A 2 F
(D) V −4 A −2 F
14. [Online January 2019]
The least count of the main scale of a screw gauge is
1 mm. The minimum number of divisions on its circular scale required to measure 5 μm diameter of a
wire is
(A) 200
(B) 50
(C) 100
(D) 500
15. [Online January 2019]
Let l , r , c and v represent inductance, resistance,
capacitance and voltage, respectively. The dimension
1
of
in SI units will be
rcv
[ A −1 ]
(C) [ LT 2 ]
(A)
(B)
[ LA −2 ]
(D) [ LTA ]
16. [2018]
The density of a material in the shape of a cube is
determined by measuring three sides of the cube and
its mass. If the relative errors in measuring the mass
and length are respectively 1.5% and 1%, the maximum error in determining the density is
(A) 2.5%
(B) 3.5%
(C) 4.5%
(D) 6%
17. [Online 2018]
In a screw gauge, 5 complete rotations of the screw
cause it to move a linear distance of 0.25 cm . There
are 100 circular scale divisions. The thickness of a
wire measured by this screw gauge gives a reading of
4 main scale divisions and 30 circular scale divisions.
Assuming negligible zero error, the thickness of the
wire is
(A) 0.3150 cm
(B) 0.2150 cm
(C) 0.4300 cm
(D) 0.0430 cm
18. [Online 2018]
The relative error in the determination of the surface
area of a sphere is α . Then the relative error in the
determination of its volume is
3
(A)
α
(B) α
2
5
2
(C)
α
(D)
α
2
3
11/28/2019 6:52:21 PM
Chapter 2: Measurements and General Physics
19. [Online 2018]
The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be
determined from a suitable combination of the fundamental physical constants G , and c . Which of the
following correctly gives the Planck length?
(A) G 2c 3
⎛ G ⎞
(C) ⎜ 3 ⎟
⎝ c ⎠
(B) G 2c
12
12 2
(D) G c
20. [Online 2018]
The percentage errors in quantities P , Q , R and S
are 0.5%, 1%, 3% and 1.5% respectively in the measureP 3Q 2
. The maximum
ment of a physical quantity A =
RS
percentage error in the value of A will be
(A) 6.5%
(B) 8.5%
(C) 6%
(D) 7.5%
21. [Online 2018]
The relative uncertainty in the period of a satellite
orbiting around the earth is 10 −2 . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mas of the earth is
(A) 6 × 10 −2
(B) 10 −2
(C) 2 × 10 −2
(D) 3 × 10 −2
22. [Online 2017]
Time ( T ) , velocity ( C ) and angular momentum ( h )
are chosen as fundamental quantities instead of mass,
length and time. In terms of these, the dimensions of
mass would be
(A) [ M ] = [ TC −2 h ]
(B) [ M ] = [ T −1C −2 h −1 ]
(C) [ M ] = [ T −1C −2 h ]
(D) [ M ] = [ T −1C 2 h ]
23. [Online 2017]
A physical quantity P is described by the relation
P = a1 2b 2c 3 d −4 . If the relative errors in the measurement of a , b , c and d respectively, are 2%, 1%, 3%
and 5%, then the relative error in P will be
(A) 25%
(B) 12%
(C) 8%
(D) 32%
24. [2016]
A student measures the time period of 100 oscillations
of a simple pendulum four times. The data set is 90 s,
91 s , 95 s and 92 s . If the minimum division in the
measuring clock is 1 s , then the reported mean time
should be
(B) 92 ± 5 s
(A) 92 ± 2 s
(C) 92 ± 1.8 s
(D) 92 ± 3 s
02_Measurements, General Physics_Part 2.indd 67
2.67
25. [Online 2016]
In the following ’I ’ refers to current and other symbols
have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity
(A) M −1L−3T 3 I
(B)
M −1L−3T 3 I 2
(C) M −1L3T 3 I
(D) ML−3T −3 I 2
26. [Online 2016]
A , B , C and D are four different physical quan
tities having different dimensions. None of them
is dimensionless. But we know that the equation
AD = C ln ( BD ) holds true. Then which of the combination is not a meaningful quantity?
(A)
C
AD2
−
BD
C
(B)
(C)
A
−C
B
(D)
A 2 − B 2C 2
(A −C)
D
27. [Online 2015]
If the capacitance of a nanocapacitor is measured in
terms of a unit u made by combining the electronic
charge e , Bohr radius a0 , Planck’s constant h and
speed of light c then
(A) u =
e 2c
ha0
(B)
u=
e2h
ca0
(C) u =
e 2 a0
hc
(D) u =
hc
e 2 a0
28. [Online 2015]
If electronic charge e , electron mass m , speed of light
in vacuum c and Planck’s constant h are taken as
fundamental quantities, the permeability of vacuum
μ0 can be expressed in units of
⎛ hc ⎞
(A) ⎜
⎝ me 2 ⎟⎠
(B)
⎛ h ⎞
⎜⎝
⎟
me 2 ⎠
⎛ h ⎞
(C) ⎜ 2 ⎟
⎝ ce ⎠
⎛ mc 2 ⎞
(D) ⎜ 2 ⎟
⎝ he ⎠
29. [Online 2015]
A beaker contains a fluid of density ρ kgm −3 , specific
heat S Jkg −1 °C −1 and viscosity η . The beaker is filled
up to height h . To estimate the rate of heat transfer per
⎛ Q ⎞
unit area ⎜ ⎟ by convection when beaker is put on
⎝ A⎠
a hot plate, a student proposes that it should depend
⎛ SΔθ ⎞
⎛ 1 ⎞
when Δθ ( in °C ) is the
and ⎜
on η , ⎜
⎝ h ⎟⎠
⎝ ρ g ⎟⎠
difference in the temperature between the bottom and
top of the fluid. In that situation the correct option for
⎛ Q ⎞
⎜⎝ ⎟⎠ is
A
11/28/2019 6:52:32 PM
2.68 JEE Advanced Physics: Mechanics – I
SΔθ
h
⎛ SΔθ ⎞ ⎛ 1 ⎞
(B) η ⎜
⎝ h ⎟⎠ ⎜⎝ ρ g ⎟⎠
SΔθ
ηh
⎛ SΔθ ⎞ ⎛ 1 ⎞
(D) ⎜
⎝ ηh ⎟⎠ ⎜⎝ ρ g ⎟⎠
(A) η
(C)
30. [2014]
A student measured the length of a rod and wrote it as
3.50 cm . Which instrument did he use to measure it?
(A)A screw gauge having 50 divisions in the circular
scale and pitch as 1 mm .
(B) A meter scale.
(C)A vernier calliper where the 10 divisions in vernier scale. Matches with 9 division in main scale
and main scale has 10 divisions in 1 cm .
(D)A screw gauge having 100 divisions in the circular scale and pitch as 1 mm .
31. [2013]
Let [ ε 0 ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time
and A = electric current, then
[ ε 0 ] = [ M −1L2T −1A ]
(B) [ ε 0 ] = [ M −1L−3T 2 A ]
(A)
[ ε 0 ] = [ M −1L−3T 4 A2 ]
(D) [ ε 0 ] = [ M −1L2T −1A −2 ]
(C)
32. [2012]
Resistance of a given wire is obtained by measuring
the current flowing in it and the voltage difference
applied across it. If the percentage errors in the measurement of the current and the voltage difference are
3% each, then error in the value of resistance of the
wire is
(A) zero
(B) 1%
(C) 3%
(D) 6%
33. [2010]
The respective number of significant figures for the
numbers 23.023, 0.0003 and 2.1 × 10 −3 are
(A) 4, 4, 2
(B) 5, 1, 2
(C) 5, 1, 5
(D) 5, 5, 2
ARCHIVE: JEE ADVANCED
Single Correct Choice Type Problems
In this section each question has four choices (A), (B), (C)
and (D), out of which ONLY ONE is correct.
1. [JEE (Advanced) 2017]
Consider an expanding sphere of instantaneous radius
R whose total mass remains constant. The expansion
is such that the instantaneous density ρ remains uniform throughout the volume. The rate of fractional
⎛ 1 dρ ⎞
change in density ⎜
is constant. The velocity v
⎝ ρ dt ⎟⎠
of any point of the surface of the expanding sphere is
proportional to
1
(B)
(A) R
R
(C) R3
2
3. [JEE (Advanced) 2016]
There are two Vernier callipers both of which have
1 cm divided into 10 equal divisions on the main scale.
The Vernier scale of one of the callipers ( C1 ) has 10
equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other calliper ( C2 ) has
10 equal divisions that correspond to 11 main scale
divisions. The readings of the two callipers are shown
in the figure. The measured values (in cm) by callipers
C1 and C2 respectively, are
(D) R 3
2. [JEE (Advanced) 2017]
A person measures the depth of a well by measuring
the time interval between dropping a stone and receiving the sound of impact with the bottom of the well.
The error in his measurement of time is δ T = 0.01 s
and he measures the depth of the well to be L = 20 m.
Take the acceleration due to gravity g = 10 ms −2 and
02_Measurements, General Physics_Part 2.indd 68
the velocity of sound is 300 ms −1 . Then the fractional
δL
, is closest to
error in the measurement,
L
(A) 1%
(B) 5%
(C) 3%
(D) 0.2%
2
3
4
C1
0
5
3
2
10
4
C2
0
5
10
11/28/2019 6:52:37 PM
2.69
Chapter 2: Measurements and General Physics
(A) 2.87 and 2.86
(C) 2.87 and 2.83
(B) 2.87 and 2.87
(D) 2.85 and 2.82
4. [JEE (Advanced) 2013]
The diameter of a cylinder is measured using a vernier
callipers with no zero error. It is found that the zero of
the vernier scale lies between 5.10 cm and 5.15 cm of
the main scale. The vernier scale has 50 division equivalent to 2.45 cm . The 24th division of the vernier scale
exactly coincides with one of the main scale divisions.
The diameter of the cylinder is
(A) 5.112 cm
(B) 5.124 cm
(C) 5.136 cm
(D) 5.148 cm
5.
[IIT-JEE 2012]
4 MLg ⎞
⎛
In the determination of Young’s modulus ⎜ Y =
⎟
⎝
π ld 2 ⎠
by using Searle’s method, a wire of length L = 2 m and
diameter d = 0.5 mm is used. For a load M = 2.5 kg ,
an extension l = 0.25 mm in the length of the wire is
observed. Quantities d and l are measured using a
screw gauge and a micrometer, respectively. They
have the same pitch of 0.5 mm . The number of divisions on their circular scale is 100. The contributions to
the maximum probable error of the Y measurement is
(A)due to the errors in the measurements of d and l
are the same.
(B)due to the error in the measurement of d is twice
that due to the error in the measurement of l .
(C)due to the error in the measurement of l is twice
that due to the error in the measurement of d .
(D)due to the error in the measurement of d is four
times that due to the error in the measurement of l.
6. [IIT-JEE 2011]
The density of a solid ball is to be determined in an
experiment. The diameter of the ball is measured with
a screw gauge, whose pitch is 0.5 mm and there are
50 divisions on the circular scale. The reading on the
main scale is 2.5 mm and that on the circular scale
is 20 divisions. If the measured mass of the ball has a
relative error of 2%, the relative percentage error in the
density is
(A) 0.9%
(B) 2.4%
(C) 3.1%
(D) 4.2%
7. [IIT-JEE 2010]
A vernier callipers has 1 mm marks on the main scale.
It has 20 equal divisions on the vernier scale which
match with 16 main scale divisions. For this vernier
callipers, the least count is
(A) 0.02 mm
(B) 0.05 mm
(C) 0.1 mm
(D) 0.2 mm
02_Measurements, General Physics_Part 2.indd 69
8. [IIT-JEE 2008]
Students I, II and III perform an experiment for measuring the acceleration due to gravity ( g ) using a
simple pendulum. They use different lengths of the
pendulum and/or record time for different number of
oscillations. The observations are shown in the table.
Least count for length = 0.1 cm , Least count for time
= 0.1 s
Total time
Length
Time
Number of
for (n)
of the
period
oscillations
Student
oscillations
pendulum
(s)
(n)
(s)
(cm)
I
64.0
8
128.0
16.0
II
64.0
4
64.0
16.0
III
20.0
4
36.0
9.0
If EI , EII and EIII are the percentage errors in g , i.e.,
⎛ Δg
⎞
⎜⎝ g × 100 ⎟⎠ for students I, II and III respectively
(A) EI = 0
(B)
(C) EI = EII
(D) EII is maximum
EI is minimum
9. [IIT-JEE 2007]
A student performs an experiment to determine the
Young’s modulus of a wire, exactly 2 m long, by
Searle’s method. In a particular reading, the student
measures the extension in the length of the wire to be
0.8 mm with an uncertainty of ±0.05 mm at a load
of exactly 1 kg . The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of
±0.01 mm . Take g = 9.8 ms −2 (exact). The young’s
modulus obtained from the reading is close to
(A) ( 2 ± 0.3 ) 1011 Nm −2
(B)
( 2 ± 0.2 ) × 1011 Nm −2
(C) ( 2 ± 0.1 ) × 1011 Nm −2
(D) ( 2 ± 0.05 ) × 1011 Nm −2
10. [IIT-JEE 2007]
In the experiment to determine the speed of sound
using a resonance column
(A)prongs of the tuning fork are kept in a vertical
plane.
(B)prongs of the tuning fork are kept in a horizontal
plane.
11/28/2019 6:52:41 PM
2.70 JEE Advanced Physics: Mechanics – I
(C)in one of the two resonances observed, the length
of the resonating air column is close to the wavelength of sound in air.
(D)in one of the two resonances observed, the length
of the resonating air column is close to half of the
wavelength of sound in air.
11. [IIT-JEE 2006]
The circular scale of a screw gauge has 50 divisions
and pitch of 0.5 mm. Find the diameter of sphere. Main
scale reading is 2.
AB
S
NH
E
′0 5
AB
S
K
NH
E
″2 25 K
R
R
15. [IIT-JEE 2004]
αZ
α − kθ
e
, P is pressure, Z is disβ
tance, k is Boltzmann constant and θ is the temperature. The dimensional formula of β will be
In the relation, P =
[ M 0 L2T 0 ]
(C) [ M 1L0T −1 ]
(A)
[ M1L2T 1 ]
(D) [ M 0 L2T −1 ]
(B)
16. [IIT-JEE 2003]
A cube has a side of length 1.2 × 10 −2 m . Calculate its
volume
(A) 1.7 × 10 −6 m 3
(B) 1.73 × 10 −6 m 3
(C) 1.70 × 10 −6 m 3
(D) 1.732 × 10 −6 m 3
17. [IIT-JEE 2001]
M
M
(A) 1.2 mm
(C) 2.20 mm
(B) 1.25 mm
(D) 2.25 mm
12. [IIT-JEE 2006]
A student performs an experiment for determination
⎛ 4π 2l ⎞
of g ⎜ = 2 ⎟ , l ≈ 1 m , and he commits an error of
⎝ T ⎠
Δl . For T he takes the time of n oscillations with
the stop watch of least count ΔT and he commits a
human error of 0.1 s. For which of the following data,
the measurement of g will be most accurate.
(A) ΔL = 0.5 , ΔT = 0.1 , n = 20
(B)
ΔL = 0.5 , ΔT = 0.1 , n = 50
(C) ΔL = 0.5 , ΔT = 0.01 , n = 20
(D) ΔL = 0.1 , ΔT = 0.05 , n = 50
13. [IIT-JEE 2005]
Which of the following sets have different dimensions?
(A) Pressure, Young’s modulus, Stress
(B) Emf, Potential difference, Electric Potential
(C) Heat, Work done, Energy
(D) Dipole moment, Electric flux, Electric field
14. [IIT-JEE 2004]
A wire has a mass 0.3 ± 0.003 g , radius 0.5 ± 0.005 mm
and length 6 ± 0.06 cm . The maximum percentage
error in the measurement of its density is
(A) 1
(B) 2
(C) 3
(D) 4
02_Measurements, General Physics_Part 2.indd 70
ΔV
, where ε 0 is the perΔt
mittivity of free space, L is a length, ΔV is a potential
difference and Δt is a time interval. The dimensional
formula for X is the same as that of
(A) resistance
(B) charge
(C) voltage
(D) current
A quantity X is given by ε 0 L
18. [IIT-JEE 2000]
⎛ 1⎞
The dimension of ⎜ ⎟ ε 0E2 ( ε 0 : permittivity of free
⎝ 2⎠
space; E : electric field) is
(A) MLT −1
(B)
ML−1T −2
(C) MLT −2
(D) ML2T −1
19. [IIT-JEE 1995]
In the formula X = 3YZ 2 , X and Z have dimensions
of capacitance and magnetic induction respectively.
What are the dimensions of Y in MKSQ system?
(A) M −3 L−1T 3Q 4
(B)
M −3 L−2T 4Q 4
(C) M −2 L−2T 4Q 4
(D) M −3 L−2T 4Q1
20. [IIT-JEE 1992]
A highly rigid cubical block A of small mass M and
side L is fixed rigidly onto another cubical block B of
the same dimensions and of low modulus of rigidity
η such that the lower face of A completely covers the
upper face of B . The lower face of B is rigidly held
on a horizontal surface. A small force F is applied
perpendicular to one of the side faces of A . After the
force is withdrawn block A executes small oscillations. The time period of which is given by
11/28/2019 6:52:48 PM
Chapter 2: Measurements and General Physics
(A) 2π
Mη
L
(B)
2π
L
Mη
(C) 2π
ML
η
(D) 2π
M
ηL
21. [IIT-JEE 1991]
If L , R , C and V respectively represent inductance
resistance, capacitance and potential difference, then
L
the dimensions of
are the same as those of
RCV
1
(A) current
(B)
current
(C) charge
(D)
1
charge
22. [IIT-JEE 1990]
If E , M , J and G respectively denote energy, mass,
angular momentum and gravitational constant, then
EJ 2
has the dimensions of
M 5G 2
(A) length
(C) mass
(B) angle
(D) time
Multiple Correct Choice Type Problems
In this section each question has four choices (A), (B), (C)
and (D), out of which ONE OR MORE is/are correct.
1. [JEE (Advanced) 2019]
Let us consider a system of units in which mass and
angular momentum are dimensionless. If length has
dimension of L, which of the following statement(s)
is/are correct?
(A) The dimension of force is L−3
(B) The dimension of power is L−5
(C) The dimension of linear momentum is L−1
(D) The dimension of energy is L−2
2. [JEE (Advanced) 2016]
A length-scale ( l ) depends on the permittivity ( ε ) of
a dielectric material. Boltzmann’s constant ( kB ) , the
absolute temperature (T), the number per unit volume
(n) of certain charged particles and the charge (q) carried by each of the particles. Which of the following
expression(s) for l is(are) dimensionally correct?
⎛ nq2 ⎞
(A) l = ⎜
⎟
⎝ ε k BT ⎠
(B)
⎛
⎞
q2
(C) l = ⎜ 2 3
⎟
⎝ ε n k BT ⎠
⎛
⎞
q2
(D) l = ⎜ 1 3
⎟
⎝ ε n k BT ⎠
02_Measurements, General Physics_Part 2.indd 71
⎛ εk T ⎞
l = ⎜ B2 ⎟
⎝ nq ⎠
2.71
3. [JEE (Advanced) 2016]
In an experiment to determine the acceleration due
to gravity g , the formula used for the time period
of a periodic motion is T = 2π
7(R − r )
. The values
5g
of R and r are measured to be ( 60 ± 1 ) mm and
( 10 ± 1 ) mm , respectively. In five successive measurements, the time period is found to be 0.52 s , 0.56 s ,
0.57 s , 0.54 s and 0.59 s . The least count of the watch
used for the measurement of time period is 0.01 s .
Which of the following statement(s) is(are) true?
(A) The error in the measurement of r is 10%
(B) The error in the measurement of T is 3.57%
(C) The error in the measurement of T is 2%
(D) The error in the determined value of g is 11%
4. [JEE (Advanced) 2015]
Planck’s constant h , speed of light c and gravitational constant G are used to from a unit of length
L and a unit of mass M . Then the correct option(s)
is(are)
(A) M ∝ c
(B)
M∝ G
(C) L ∝ h
(D) L ∝ G
5. [JEE (Advanced) 2015]
In terms of potential difference V , electric current I ,
permittivity ε 0 , permeability μ0 and speed of light c,
the dimensionally correct equation(s) is(are)
(A) μ0 I 2 = ε 0V 2
(B)
(C) I = ε 0cV
(D) μ0cI = ε 0V
ε 0 I = μ0V
6. [JEE (Advanced) 2015]
Consider a vernier calliper in which each 1 cm on
the main scale is divided into 8 equal divisions and
a screw gauge with 100 divisions on its circular scale.
In the vernier callipers, 5 divisions of the Vernier scale
coincide with 4 divisions on the main scale and in
the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale.
Then
(A)if the pitch of the screw gauge is twice the least
count of the Vernier callipers, the least count of
the screw gauge is 0.01 mm
(B)if the pitch of the screw gauge is twice the least
count of the vernier calliper, the least count of the
screw gauge is 0.05 mm
(C)if the least count of the linear scale of the screw
gauge is twice the least count of the vernier callipers, the least count of the screw gauge is 0.01 mm.
(D)if the least count of the linear scale of the screw
gauge is twice the least count of the Vernier calliper, the least count of the screw gauge is 0.005 mm.
11/28/2019 6:52:54 PM
2.72 JEE Advanced Physics: Mechanics – I
7. [JEE (Advanced) 2010]
A student uses a simple pendulum of exactly 1 m
length to determine g , the acceleration due to gravity. He uses a stop watch with the least count of 1 s
for this and records 40 s for 20 oscillations. For this
observation, which of the following statement(s) is/
are true?
(A)Error ΔT in measuring T , the time period, is
0.05 s
(B) Error ΔT in measuring T , the time period, is 1 s
(C) Percentage error in the determination of g is 5%
(D) Percentage error in the determination of g is 2.5%
8. [IIT-JEE 1998]
Let [ ε 0 ] denotes the dimensional formula of the
permittivity of the vacuum and [ μ0 ] that of the permeability of the vacuum. If M = mass , L = length ,
T = Time and I = electric current , then
(A)
[ ε 0 ] = M −1L−3T 2I
(B)
[ ε 0 ] = M −1L−3T 4 I 2
(C)
[ μ0 ] = MLT −2I −2
(D)
[ μ0 ] = ML2T −1I
9. [IIT-JEE 1998]
The SI unit of the inductance, the henry can by written
as
weber
volt-second
(B)
(A)
ampere
ampere
(C)
joule
( ampere )2
(D) ohm-second
10. [IIT-JEE 1995]
The pairs of physical quantities that possess same
dimensions is/are
(A) Reynold’s number and coefficient of friction
(B) Curie and frequency of light wave
(C) Latent heat and gravitational potential
(D) Planck’s constant and torque
(C) Energy and Young’s modulus
(D) Light year and wavelength
13. [IIT-JEE 1984]
L , C and R represent the physical quantities induc
tance, capacitance and resistance respectively. The
combinations which have the dimensions of frequency
are
1
1
, y=
2
2
1
3
(C) y = , z =
2
2
(B)
x=
1
1
, z=
2
2
3
1
(D) y = − , z =
2
2
12. [IIT-JEE 1986]
The pair(s) having same dimensions is/are
(A) Torque and work
(B) Angular momentum and work
02_Measurements, General Physics_Part 2.indd 72
1
RC
(B)
R
L
(C)
1
LC
(D)
C
L
Linked Comprehension Type Problems
This section contains Linked Comprehension Type
Questions or Paragraph based Questions. Each set consists
of a Paragraph followed by questions. Each question has
four choices (A), (B), (C) and (D), out of which only one is
correct. (For the sake of competitiveness there may be a few
questions that may have more than one correct options)
Comprehension 1
[JEE (Advanced) 2018]
In electromagnetic theory, the electric and magnetic phenomena are related to each other. Therefore, the dimensions
of electric and magnetic quantities must also be related to
each other. In the questions below, [ E ] and [ B ] stand for
dimensions of electric and magnetic fields respectively,
while [ ε 0 ] and [ μ0 ] stand for dimensions of the permittivity and permeability of free space, respectively. [ L ] and
[ T ] are dimensions of length and time, respectively. All the
quantities are given in SI units.
(There are two questions based on above paragraph,
the question given below is one of them).
1.
The relation between [ E ] and [ B ] is
(A) [ E ] = [ B ][ L ][ T ]
11. [IIT-JEE 1992]
If the dimensions of length are expressed as G x c y h z ;
where G , c and h are the universal gravitational
constant, speed of light and Planck’s constant respectively, then
(A) x =
(A)
−1
(B) [ E ] = [ B ][ L ] [ T ]
−1
(C) [ E ] = [ B ][ L ][ T ]
−1
−1
(D) [ E ] = [ B ][ L ] [ T ]
2.
The relation between [ ε 0 ] and [ μ0 ] is
(A)
[ μ0 ] = [ ε 0 ][ L ]2 [ T ]−2
(B)
[ μ0 ] = [ ε 0 ][ L ]−2 [ T ]2
(C)
[ μ0 ] = [ ε 0 ]−1 [ L ]2 [ T ]−2
(D)
[ μ0 ] = [ ε 0 ]−1 [ L ]−2 [ T ]2
11/28/2019 6:53:01 PM
Chapter 2: Measurements and General Physics
2.73
Comprehension 2
Comprehension 3
[JEE (Advanced) 2018]
If the measurement errors in all the independent quantities are known, then it is possible to determine the error in
any dependent quantity. This is done by the use of series
expansion and truncating the expansion at the first power
x
of the error. For example, consider the relation z = . If the
y
[JEE (Advanced) 2011]
A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be
treated as neutral plasma. Let N be the number density of
free electrons, each of mass m . When the electrons are subjected to an electric field, they are displaced relatively away
from the heavy positive ions. If the electric field becomes
zero, the electrons begin to oscillate about the positive ions
with a natural angular frequency ω p , which is called the
plasma frequency.
To sustain the oscillations, a time varying electric
field needs to be applied that has an angular frequency w,
where a part of the energy is absorbed and a part of it is
reflected. As ω approaches ω p , all the free electrons are set
to resonance together and all the energy is reflected. This is
the explanation of high reflectivity of metals.
errors in x , y and z are Δx , Δy and Δz respectively,
then z ± Δz =
Δy ⎞
x ± Δx x ⎛
Δx ⎞ ⎛
= ⎜ 1±
⎟⎠ ⎜ 1 ±
⎝
y ± Δy y
x ⎝
y ⎟⎠
−1
.
−1
Δy ⎞
⎛
The series expansion for ⎜ 1 ±
, to first power in
y ⎠⎟
⎝
Δy
⎛ Δy ⎞
, is 1 ∓ ⎜
. The relative errors in independent variy
⎝ y ⎟⎠
ables are always added. So, the error in z will be
5.
⎛ Δx Δy ⎞
Δz = z ⎜
+
y ⎟⎠
⎝ x
Δx
The above derivation makes the assumption that
1,
x
Δy
1 . Therefore, the higher powers of these quantities
y
are neglected.
(There are two questions based on above paragraph,
the question given below is one of them)
3.
(1 − a )
to be determined by
Consider the ratio r =
(1 + a )
measuring a dimensionless quantity a . If the error in
⎛ Δa
⎞
1 ⎟ , then what is
the measurement of a is Δa ⎜
⎝ a
⎠
the error Δr in determining r ?
(A)
(C)
4.
Δa
( 1 + a )2
2 Δa
( 1 − a )2
(B)
(D)
2 Δa
( 1 + a )2
2 aΔa
( 1 − a2 )
In an experiment, the initial number of radioactive nuclei is 3000. It is found that 1000 ± 40 nuclei
decayed in the first 1 s . For x 1 , ln ( 1 + x ) = x up
to first power in x. The error Δλ , in the determination
of the decay constant λ in s −1 , is
(A) 0.04
(B) 0.03
(C) 0.02
(D) 0.01
02_Measurements, General Physics_Part 2.indd 73
6.
Taking the electronic charge as e and the permittivity
as ε 0 , use dimensional analysis to determine the correct expression for ω p
(A)
Ne
mε 0
(B)
mε 0
Ne
(C)
Ne 2
mε 0
(D)
mε 0
Ne 2
Estimate the wavelength at which plasma reflection
will occur for a metal having the density of electrons
N ≈ 4 × 10 27 m −3. Take ε 0 ≈ 10 −11 and m ≈ 10 −30 , where
these quantities are in proper SI units
(A) 800 nm
(B) 600 nm
(C) 300 nm
(D) 200 nm
Matrix Match/Column Match Type Questions
Each question in this section contains statements given in
two columns, which have to be matched. The statements
in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any
given statement in COLUMN-I can have correct matching
with ONE OR MORE statement(s) in COLUMN-II. The
appropriate bubbles corresponding to the answers to these
questions have to be darkened as illustrated in the following examples:
If the correct matches are A → p, s and t; B → q and r;
C → p and q; and D → s and t; then the correct darkening of
bubbles will look like the following:
11/28/2019 6:53:06 PM
2.74 JEE Advanced Physics: Mechanics – I
A
B
C
D
p
p
p
p
p
q
q
q
q
q
r
s
t
r
r
r
r
s
s
s
s
t
t
t
t
1. [IIT-JEE 2009]
COLUMN-II gives certain systems undergoing a
process. COLUMN-I suggests changes in some of
the parameters related to the system. Match the statements in COLUMN-I to the appropriate process(es)
from COLUMN-II.
COLUMN-I
COLUMN-II
(A) The energy of
the system is
increased.
(p) System: A capacitor,
initially uncharged.
Process: It is connected
to a battery.
(B) Mechanical
(q) System: A gas in an
energy is
adiabatic container
provided to the
fitted with an adiabatic
system, which
piston.
is converted
Process: The gas is
into energy
compressed by pushing
of random
the piston.
motion of its
parts.
(C) Internal
energy of
the system is
converted into
its mechanical
energy.
(r) System: A gas in a rigid
container.
Process: The gas
gets cooled due to
colder atmosphere
surrounding it.
(D) Mass of the
system is
decreased.
(s) System: A heavy
nucleus, initially at rest.
Process: The nucleus
fissions into two
fragments of nearly
equal masses and some
neutrons are emitted.
(t) System: A resistive wire
loop.
Process: The loop
is placed in a time
varying magnetic field
perpendicular to its
plane.
02_Measurements, General Physics_Part 2.indd 74
2. [IIT-JEE 2009]
COLUMN-II shows five systems in which two objects
are labelled as X and Y . Also in each case a point P
is shown. COLUMN-I gives some statements about
X and/or Y . Match these statements to the appropriate system(s) from COLUMN-II.
COLUMN-I
COLUMN-II
(A) The force
exerted by
X on Y has a
magnitude
Mg.
(p) Block Y of mass M
left on a fixed inclined
plane X, slides on it
with a constant velocity.
Y
X
P
(B) The
gravitational
potential
energy of X is
continuously
increasing.
(q) Two ring magnets Y
and Z, each of mass M,
are kept in frictionless
vertical plastic stand
so that they repel each
other. Y rests on the
base X and Z hangs in
air in equilibrium. P is
the topmost point of the
stand on the common
axis of the two rings.
The whole system is in a
lift that is going up with
a constant velocity.
P
Z
Y
X
(Continued)
11/28/2019 6:53:07 PM
2.75
Chapter 2: Measurements and General Physics
COLUMN-I
COLUMN-II
(C) Mechanical
energy of
the system
X + Y is
continuously
decreasing.
(r) A pulley Y of mass m0
is fixed to a table through
a clamp X. A block of mass
M hangs from a string that
goes over the pulley and is
fixed at point P of the table.
The whole system is kept
in a lift that is going down
with a constant velocity.
P
Y
3. [IIT-JEE 2007]
Some physical quantities are given in COLUMN-I and
some possible SI units in which these quantities may
be expressed are given in COLUMN-II. Match the
physical quantities in COLUMN-I with the units in
COLUMN-II.
COLUMN-I
(p) (volt)
(A) GMeMs
(coulomb)
G – universal gravitational
(metre)
constant,
Me – mass of the earth,
Ms – mass of the sun.
(B)
X
(D) The torque of
the weight of
Y about point
P is zero.
3RT
M
R – universal gas constant,
T – absolute temperature,
M – molar mass.
(s) A sphere Y of mass M
is put in a non-viscous
liquid X kept in a
container at rest. The
sphere is released and
it moves down in the
liquid.
(C)
F2
q 2B2
F – force,
q – charge,
B – magnetic field.
(D)
GMe
Re
G – universal gravitational
constant,
Me – mass of the earth,
Re – radius of the earth.
Y
X
P
(t) A sphere Y of mass M is
falling with its terminal
velocity in a viscous
liquid X kept in a
container.
Y
X
COLUMN-II
(q) (kilogram)
( metre )3
( second )−2
2
(r) ( metre )
( second )−2
(s) (farad)
( volt )2
( kg )−1
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical
value obtained after series of calculations based on the data
provided in the question(s).
1. [JEE (Advanced) 2019]
An optical bench has 1.5 m long scale having four
equal divisions in each cm . While measuring the focal
length of a convex lens, the lens is kept at 75 cm mark
of the scale and the object pin is kept at 45 cm mark.
The image of the object pin on the other side of the
lens overlaps with image pin that is kept at 135 cm
mark. In this experiment, the percentage error in the
measurement of the focal length of the lens is ……….
P
02_Measurements, General Physics_Part 2.indd 75
11/28/2019 6:53:10 PM
2.76 JEE Advanced Physics: Mechanics – I
2. [JEE (Advanced) 2018]
A steel wire of diameter 0.5 mm and Young’s modulus 2 × 1011 Nm −2 carries a load of mass m . The length
of the wire with the load is 1 m . A vernier scale with
10 divisions is attached to the end of this wire. Next to
the steel wire is a reference wire to which a main scale,
of least count 1 mm , is attached. The 10 divisions of
the vernier scale correspond to 9 divisions of the main
scale. Initially, the zero of vernier scale coincides with
the zero of main scale. If the load on the steel wire is
increased by 1.2 kg , the vernier scale division which
coincides with a main scale division is ………. (Take,
g = 10 ms −2 and π = 3.2 )
3. [JEE (Advanced) 2015]
The energy of a system as a function of time t is given
as E ( t ) = A 2exp ( −α t ) , where α = 0.2 s −1 . The measurement of A has an error of 1.25%. If the error in the
measurement of time is 1.50%, the percentage error in
the value of E ( t ) at t = 5 s is
02_Measurements, General Physics_Part 2.indd 76
4. [JEE (Advanced) 2014]
To find the distance d over which a signal can be seen
clearly in foggy conditions, a railway engineer uses
dimensional analysis and assumes that the distance
depends on the mass density r of the fog, intensity
(power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to
S1 n . The value of n is
5. [JEE (Advanced) 2014]
During Searle’s experiment, zero of the vernier scale
lies between 3.20 × 10-2 m and 3.25 × 10-2 m of the main
scale. The 20th division of the vernier scale exactly
coincides with one of the main scale divisions. When
an additional load of 2 kg is applied to the wire, the
zero of the vernier scale still lies between 3.20 × 10-2 m
and 3.25 × 10-2 m of the main scale but now the 45th
division of vernier scale coincides with one of the main
scale divisions. The length of the thin metallic wire is
2 m and its cross-sectional area is 8 × 10-7 m2. The least
count of the vernier scale is 1 × 10-5 m. The maximum
percentage error in the Young’s modulus of the wire is
11/28/2019 6:53:12 PM
Chapter 2: Measurements and General Physics
2.77
Answer Keys—Test Your Concepts and Practice Exercises
Test Your Concepts-I (Based on Principle of
Homogeneity: Verification)
1. 1 in mass, 0 in length, −2 in time
2. Yes
6. M 0 L0T 0
7. m −3 L−2T 4Q 4
8. True
9. A : no units, B : m2
3. k
m
f
W
4. k ⎛⎜ ⎞⎟
⎝ ηr ⎠
⎛ r4 ⎞
5. k ⎜ ⎟ p
⎝ η⎠
η
6. k ⎛ 2 ⎞
⎜⎝ ρd ⎟⎠
v
10. No, tan θ =
u
Test Your Concepts-II (Based on Principle of
Homogeneity: Conversion)
1. 4.18 α −1β −2γ 2 new units
S ⎞
7. k ⎛
⎜⎝ rgd ⎟⎠
9. ρ g 4 ν 6
10. (a) Kv −1FT
(b) KvFT
2. 109 erg
3. 10 poise
4. 8 kg
5. 50 new units
6. 746 W
Test Your Concepts-IV
(Based on Errors, Significant Figures, Vernier
Calliper and Screw Gauge)
ρv 3
ν
8. 500 new units of length
7. p ∝ ρv 4ν 3 and σ ∝
1. ( 4 ± 0.65 ) Ω , 4 Ω ± 16.25%
2. 8.1 gcc −1 ± 3.6%
9. 10 4 kg
1
cm
10N
10. 1.4 kWm −2
3.
Test Your Concepts-III (Based on Principle of
Homogeneity: Dependence)
4. 43.7 cm 3
5. 5, 3, 2, 3, 5, 2, 2
6. 7.36, 8.33, 9.44, 15.8, 7.37, 9.44, 15.8
7. (a) 11.4; (b) 2.4; (c) 44; (d) 108
8. (a) 0.01 cm; (b) 10.23 cm
9. 6.4 mm
10. Actual length will be 0.05 cm less than the measured
length.
1. T = k
−
5 1
r 3d
S
1
2. kp 6 d 2 E 3
Single Correct Choice Type Questions
1. D
2. A
3. B
4. C
5. B
6. C
7. A
8. A
9. A
10. A
11. A
12. B
13. A
14. B
15. A
16. B
17. C
18. D
19. D
20. C
21. B
22. D
23. C
24. A
25. B
26. B
27. A
28. D
29. D
30. B
31. B
32. B
33. B
34. C
35. C
36. C
37. C
38. B
39. A
40. B
41. D
42. B
43. A
44. D
45. A
46. A
47. C
48. D
49. D
50. B
51. A
52. C
53. A
54. D
55. A
56. D
57. C
58. C
59. B
60. C
02_Measurements, General Physics_Part 2.indd 77
11/28/2019 6:53:15 PM
2.78 JEE Advanced Physics: Mechanics – I
61. C
62. B
63. B
64. C
65. A
66. A
67. B
68. B
69. D
70. C
71. B
72. D
73. C
74. A
75. D
76. D
77. A
78. A
79. B
80. A
81. C
82. C
83. A
84. B
85. A
86. C
87. B
88. D
89. B
90. C
91. A
92. B
93. A
94. B
95. B
96. D
97. A
98. D
99. B
100. A
101. C
102. D
103. C
104. C
105. C
106. B
107. B
108. D
109. B
110. C
111. C
112. D
113. B
114. D
115. C
116. D
117. A
118. B
119. C
120. C
121. B
122. A
123. C
124. B
125. A
126. A
127. B
128. D
129. B
130. C
131. A
132. A
133. C
134. A
135. D
136. A
137. A
138. C
139. B
140. A
141. A
142. B
143. B
144. B
145. C
146. C
147. C
148. C
149. A
150. D
151. D
152. A
153. C
154. B
155. A
156. A
157. C
158. A
159. A
160. C
161. D
Multiple Correct Choice Type Questions
1. A, C
2. A, B, C
3. A, B, C, D
4. A, B, D
5. C
6. A, B, C
7. A, D
8. A, B, C
9. B, C
10. A, B, C
11. A, B, D
12. A, B, C
13. B, C
14. A, B, C
15. A, B, D
16. B, C, D
17. A, B
18. A, B, D
19. A, B, C
20. B, C, D
21. A, B, C, D
22. A, B, C
23. A, B, D
24. B, C
25. A, C
26. A, B, C
27. B, C, D
28. B, C, D
29. A, B, D
30. A, B, C, D
Reasoning Based Questions
1. A
2. B
3. D
4. A
5. A
6. B
7. D
8. A
9. A
10. D
11. C
12. D
13. A
14. C
15. B
16. C
17. C
18. D
19. C
20. A
21. B
22. A
23. A
24. A
25. A
Linked Comprehension Type Questions
1. A, C
2. B
3. C
4. B
5. A
6. B
7. C
8. C
9. D
10. A
11. A
12. B
13. C
14. B
15. D
16. C
17. C
18. A
19. A, D
20. A, B, C
21. A, C
22. B
23. C
24. A
25. A
26. B
27. D
28. A
29. C
30. B
31. B
32. D
33. A
34. A
35. B
36. C
37. D
38. A
39. C
40. B
41. D
42. C
43. C
44. A
45. B
46. B
Matrix Match/Column Match Type Questions
1. A → (q)
B → (p)
C → (r)
D → (t)
2. A → (p, q)
B → (r, t)
C → (p, q)
D → (s)
3. A → (s)
B → (p)
C → (r)
D → (q)
4. A → (q, r)
B → (q)
C → (p, s, t)
D → (q)
5. A → (r, s)
B → (p, q)
C → (t)
D → (p, q)
6. A → (p, t)
B → (r)
C → (q)
D → (r)
7. A → (r)
B → (s)
C → (p, q, r)
D → (t)
02_Measurements, General Physics_Part 2.indd 78
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2.79
Chapter 2: Measurements and General Physics
8. A → (p, q)
B → (r, t)
C → (p, q)
D → (s)
9. A → (p, q)
B → (r, s)
C → (r, s)
D → (r, s)
10. A → (r)
B → (p)
C → (p)
D → (q)
11. A → (r)
B → (p, q)
C → (p)
D → (s)
12. A → (q, s)
B → (r)
C → (p)
D → (r)
13. A → (s)
B → (p)
C → (q)
D → (r)
14. A → (q)
B → (p)
C → (s)
D → (r)
15. A → (r)
B → (p)
C → (p)
D → (q)
16. A → (q)
B → (r)
C → (s)
D → (p)
17. A → (r)
B → (p, q)
C → (p)
D → (s)
18. A → (q)
B → (p)
C → (s)
D → (r)
19. A → (q)
B → (p)
C → (r)
D → (s)
Integer/Numerical Answer Type Questions
1. a = 0 , b = 0 , c = 0
2. x = 2
3. x = 4 , y = 1 , z = 1
4. 8 kg , 10 m and 2 s
5. x = 1 , y = 1 , z = 2
6. a = 2 , b = 3 , c = 1 , d = 1
ARCHIVE: JEE MAIN
1. D
2. D
3. A
4. C
5. B
6. D
7. C
8. C
9. A
10. C
11. C
12. B
13. C
14. A
15. A
16. C
17. B
18. A
19. C
20. A
21. C
22. C
23. D
24. A
25. B
26. A, D
27. C
28. C
29. A
30. C
31. C
32. D
33. B
ARCHIVE: JEE advanced
Single Correct Choice Type Problems
1. A
2. A
3. C
4. B
5. C
6. C
7. D
8. B
9. B
10. A
11. A
12. D
13. D
14. D
15. A
16. A
17. D
18. B
19. B
20. D
21. B
22. B
Multiple Correct Choice Type Problems
1. A, C, D
2. B, D
3. A, B, D
4. A, C, D
5. A, C
6. B, C
7. A, C
8. B, C
9. A, B, C, D
10. A, B, C
11. B, D
12. A, D
13. A, B, C
Linked Comprehension Type Questions
1. C
2. D
02_Measurements, General Physics_Part 2.indd 79
3. B
4. C
5. C
6. B
11/28/2019 6:53:17 PM
2.80 JEE Advanced Physics: Mechanics – I
Matrix Match/Column Match Type Questions
1. A → (p, q, s, t)
B → (q)
C → (s)
D → (s)
2. A → (p, t)
B → (q, s, t)
C → (p, r, t)
D → (q)
3. A → (p, q)
B → (r, s)
C → (r, s)
D → (r, s)
Integer/Numerical Answer Type Questions
1. 1.39
02_Measurements, General Physics_Part 2.indd 80
2. 3
3. ± 4%
4. 3
5. 4
11/28/2019 6:53:17 PM
CHAPTER
3
Vectors
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Vector Fundamentals and Types
(d) Vector Resolution
(b) Triangle, Parallelogram and Polygon Law
(e) Dot Product and Cross Product
of Vector Addition
(f) Scalar Triple Product and Vector Triple
(c) Position Vector and Vector Subtraction
Product
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
and Advanced) are also given.
intRoDuCtion
Physical quantities having magnitude, direction and
obeying laws of vector algebra are called vectors.
Magnitude of a vector is also called its modulus.
ExamplE:
Displacement, velocity, acceleration, momentum, force,
impulse, weight, thrust, torque, angular momentum,
angular velocity.
Outward
Normal
Conceptual Note(s)
(a) “If you love hot pizzas, then it does not imply that
everything hot is a pizza”.
(b) Similarly if a physical quantity has magnitude and
direction both, then it does not always imply that
it is a vector. For it to be a vector the third condition of obeying laws of vector algebra has to be
satisfied.
Example: The physical quantity CURRENT has both
magnitude and direction but is still a scalar as it disobeys the laws of vector algebra.
illuStRAtion 1
The physical quantity ‘AREA’ is a vector with its
direction along the outward normal to the surface.
03_Vectors_Part 1.indd 1
We can order events in time, and there is a sense
of time, distinguishing past, present and future. Is
therefore time a vector?
11/28/2019 6:55:55 PM
3.2 JEE Advanced Physics: Mechanics – I
TYPES OF VECTOR
Solution
Time always flow from past to present and then to
future so a direction can be assigned to time. However
as its direction is unique it is not to be specified. As
direction is not to be specified so time cannot be a
vector though it has direction. Also it does not obey
laws of vector algebra.
GEOMETRICAL DEFINITION
A
line segment is called a vector, denoted as
directed
AB
(read
as
AB vector). In printing we can denote
AB as AB (bold letters). Even if you have to write
BA ­vector the arrow head
above would always point
from left to right i.e. BA .
Directed line
segment
Support
A
Terminating
point/Head
Also note that if we have to compare vectors then we
only compare their magnitudes. So,
A > B and not A > B
Conceptual Note(s)
( AB )
has the following three
(a)Length: The length of AB is denoted by AB or
simply AB.
(b) Support: The line of unlimited length of which
AB is segment is called the support of vector AB.
(c) Sense:
The sense of AB is from A → B and that
of BA will be from B → A. The sense of directed
line segment is from its initial point to the terminal point.
03_Vectors_Part 1.indd 2
Problem Solving Technique(s)
A vector can always be transported parallel to its
­original direction.
Two vectors A and B are said to be parallel when
But mathematically AB = −BA
Also, let A and B be two vectors,
then A = A = magnitude of A = length of A
B = B = magnitude of B = length of B
Every vector
characteristics
Two vectors A and B are said to be equal when
they have equal magnitudes and same direction.
Geometrically if head of one vector coincides will the
head of other and so do the tails coincide then the
vectors are said to be equal.
If A = B , then A = B , always.
But if, A = B doesn’t always imply A = B.
Parallel Vector
B
Initial/Start
point tail
Equal Vectors
(a) both have same direction.
(b) magnitude of one is scalar multiple of magnitude
of the other
Example:
B = 2 A, i.e., Magnitude of B is twice the magnitude of A
and both have same direction.
A
B = 2A
A
B = –2A
In general if A = kB (where k is any constant) i.e.
magnitude of A is k times that of B , then both are
parallel if k > 0.
Anti-parallel Vectors
Two vectors A and B are said to be antiparallel when
(a) both have opposite direction.
(b) magnitude of one is a scalar multiple of the
­magnitude of the other.
Example:
B = −2 A Magnitude of B is twice the magnitude of A
and both have opposite direction.
In general if A = kB (where k is any constant) i.e.
magnitude of A is k times that of B , then both are
antiparallel if k > 0.
11/28/2019 6:56:01 PM
Chapter 3: Vectors
Collinear Vectors
When the vectors under consideration can share the
same support or have a common support then the
considered vectors are collinear.
A
3.3
Conceptual Note(s)
No units are to be attached (like newton (N), msec −1,
metre etc.) with a unit vector i.e., unit vector is dimensionless physical quantity.
B = 2A
Orthogonal Unit Vectors/Base Vectors
C = –3A
î , ĵ and k̂ are called orthogonal unit vectors. These
vectors must form a Right Handed Triad. It is a coordinate system such that when we Curl the fingers of
right hand from x to y , then we must get the direction of z along thumb
Common support
shared by A, B and C
Zero Vector ( 0 )
y
A vector having zero magnitude and arbitrary direction (not known to us) is a zero vector.
Properties of 0
j
(a) A + ( − A ) = 0
(b) If A + B = 0 ⇒ A = −B
(c) k 0 = 0 [read as k times 0 equals 0 ]
{where, k is any scalar}
(d) 0 A = 0 [read as zero times A equals zero
vector]
(e) A + 0 = 0 + A = A
x
i
z
from y to z , then we must get the direction of x
along thumb
from z to x , then we must get the direction of y
along thumb
i = x
y+
x
(2, 3)
⇒ x = xi
3j
j = y
–5i
O
–
y
x
x+
6
2i
2
–j
j
⇒ y = y j
i–
(–5, –1) –5
z
k =
z
y–
Graphical
representation
of vectors
⇒ z = zk
= 1, B
= 1, C
= 1 , x = 1 , y = 1
A
=A
Since, A
A
⇒ A = AA
B = B B
Vector has Magnitude Direction
two essential
ingredients
Thus, we conclude that unit vector gives us the
direction.
03_Vectors_Part 1.indd 3
2i
13
A vector dividedby its magnitude is a unit vector.
(read as A cap/A carat/A
Unit vector for A is A
.
hat). Unit vector for B is B
+3
j
Unit Vector
k
Example:
(a) A vector of 3 unit along x axis is x = 3i.
(b) A vector of magnitude 6 along -x axis is
x = 6 ( −i ) = −6i.
(c) A vector of magnitude 5 along -z axis is
x = 5 ( −
k ) = −5
k.
Also, ( 0, 0 ) ≡ 0i + 0j ≡ 0 , ( 2, 3 ) ≡ 2i + 3j
( −5, − 1) ≡ −5i − j.
11/28/2019 6:56:08 PM
3.4 JEE Advanced Physics: Mechanics – I
Fixed Vector
Co-terminus Vectors
Fixed vector is that vector whose initial point or tail is
fixed. Its also called localised vector.
Co-terminus vectors are those vectors which have
common terminating point.
Example:
(a) Position vector
(b) Displacement vector
c
b
d
Free Vector
a
Free vector is that vector whose initial point or tail is
not fixed. Its also known as non localised vector.
Example:
Velocity vector of a particular moving along a straight
line is a free vector.
Vectors are said to be coplanar if they occur in same
or common plane. e.g. A and B and C . Remember
that any two vectors always lie in the same plane.
Polar Vector
d
Vectors producing straight line linear effect are
called polar vectors e.g. force, momentum, velocity,
displacement.
c
b
O
Coplanar Vector
a
Axial Vectors/Pseudo Vectors:
(Virtual, Imaginary)
Negative Vector
A vector is said to be negative of a given vector if its
magnitude is the same as that of the given vector but
direction is reversed. e.g. The negative vector of a
­vector a is denoted by −a .
b (= –a )
a
If b is negative of a , then b = − a.
Co-initial Vectors
The rotational effect of a polar vector gives rise to a
new vector called axial vector (­acting along axis of
rotation).
Example:
Rotational effect of force (polar vector) is torque (an axial
vector).
Rotational effect of linear momentum (a polar v­ ector) is
angular momentum (an axial vector).
Axial vector
Axis of rotation
Anti Clockwise
Clockwise rotation
Co-initial vectors are those vectors which have the
same initial point.
a
d
b
Axis of rotation
Axial vector
c
03_Vectors_Part 1.indd 4
11/28/2019 6:56:10 PM
Chapter 3: Vectors
Conceptual Note(s)
(a) Whenever angle between two vectors is to be
taken we must make sure that either their heads
coincide or their tails coincide.
B
180° – θ
–A
θ
θ
Triangle Law of Vector Addition
of Two Vectors
If two non zero vectors are represented by the two
sides of a triangle taken in same order then the resultant is given by the closing side of triangle in opposite
order. i.e.,
R = A+B
A
B
180° – θ
O A + A B = OB
R=
–B
O
i.e., if two vectors have their heads coinciding or
their tails coinciding, then internal angle is the
angle between two vectors (whether acute or
obtuse), as in (1), (2), (5) & (6).
If head coincides with tail then external supplementary angle (whether acute or obtuse) is the
angle between the two vectors, as in (3) & (4).
θ
(2)
(4)
A
A
R
β
O
B sin θ
B
θ
A
A
N
θ
(5)
In ΔABN,
θ
⇒
(6)
Illustration 2
Let A = 2iˆ + ˆj , B = 3 ˆj − kˆ and C = 6iˆ − 2kˆ , then find
A − 2B + 3C . Also calculate its magnitude.
iˆ A − 2B + 3C = (2iˆ + ˆj ) − 2(3 ˆj − kˆ ) + 3(6iˆ − 2kˆ )
A − 2B + 3C = 2iˆ + ˆj − 6 ˆj + 2kˆ + 18iˆ − 6 kˆ
2
2
2
A − 2B + 3C = ( 20 ) + ( −5 ) + ( −4 ) = 441 = 21
03_Vectors_Part 1.indd 5
B
Magnitude of Resultant Vector
(3)
(b) If angle between A and B is θ, then
(i) angle between −A and B is ( 180° − θ )
(ii) angle between A and −B is ( 180° − θ )
(iii) angle between −A and −B is θ
Solution
B
B cos θ
θ
θ
A+
B
θ
(1)
3.5
cos θ =
AN
B
AN = B cos θ
⇒
sin θ =
BN
B
BN = B sin θ
In ΔOBN , we have OB2 = ON 2 + BN 2
2
2
⇒
R2 = ( A + B cos θ ) + ( B sin θ ) ⇒
R2 = A 2 + B2 cos 2 θ + 2 AB cos θ + B2 sin 2 θ ⇒
R2 = A 2 + B2 ( cos 2 θ + sin 2 θ ) + 2 AB cos θ ⇒
R2 = A 2 + B2 + 2 AB cos θ ⇒
R = A 2 + B2 + 2 AB cos θ Direction of Resultant Vector
If θ is angle between A and B , then
B + A = A 2 + B2 + 2 AB cos θ .
11/28/2019 6:56:16 PM
3.6 JEE Advanced Physics: Mechanics – I
If the resultant, R makes an angle b with A, then in
ΔOBN , then
tan β =
BN
BN
=
ON OA + AN
tan β =
B sin θ
A + B cos θ
Solution
METHOD I
Let F1 and F2 be the two forces. Let one of them, F1
(say) is of smaller magnitude.
Now F1 + F2 = 18 …(1)
R = F12 + F22 + 2 F1 F2 cos θ = 12 …(2)
Conceptual Note(s)
Resultant R of two vectors A and B inclined at an
angle q is
R = R = A + B = A2 + B2 + 2 AB cosθ
and β is the angle made by the resultant R with A ,
B sinθ
then tan β =
A + B cosθ
Illustration 3
Can the magnitude of the resultant of two equal vectors be equal to the magnitude of either of the ­vectors?
Explain your answer.
Solution
Let the two vectors be A and B inclined at an angle
θ , then
tan α =
F2 sin θ
→∞
F1 + F2 cos θ
{∵ α = 90° }
⇒
F1 + F2 cos θ = 0 …(3)
⇒
( 18 − F2 ) + F2 cos θ = 0 ⇒
F2 ( 1 − cos θ ) = 18 …(4)
Subtracting equation (2) from equation (1), we get
2
2
2 F1 F2 ( 1 − cos θ ) = ( 18 ) − ( 12 ) = 180 …(5)
Dividing (5) by (4), we have
2 F1 = 10
⇒
F1 = 5 N and F2 = 13 N METHOD II
Since, ( F1 + F2 ) is resultant of F1 and F2 , so from
Triangle Law, we have drawn the diagram from
which we get
R2 = A 2 + B2 + 2 AB cos θ
According to the question, we have been given
F2
R=A=B
⇒
A 2 = A 2 + A 2 + 2 A 2 cos θ ⇒
cos θ = −
⇒ θ = 120° 1
2
F1 + F2 = 12
F1
F22 = F12 + ( 12 )
2
F22 − F12 = 144 So, this is possible, when two vectors of equal magnitude are inclined to each other at an angle of 120° .
⇒
Illustration 4
Since, F1 + F2 = 18 …(2)
The sum of magnitude of two forces (in newton)
acting at a point is 18 and the magnitude of their
resultant (in newton) is 12. If resultant is at 90° with
force of smaller magnitude, what are magnitudes of
forces?
03_Vectors_Part 1.indd 6
⇒
( F1 + F2 ) ( F2 − F1 ) = 144 …(1)
So, (1) gives F2 − F1 = 8 …(3)
Solving (2) and (3), we get
F1 = 5 N and F2 = 13 N
11/28/2019 6:56:23 PM
Chapter 3: Vectors
Illustration 5
The x and y components of vector A are 4 m and
6 m respectively. The x and y components of vector
A + B are 10 m and 9 m respectively. Calculate the
(a) x and y components of vector B.
(b) length of vector B .
(c) angle which B makes with x-axis.
Magnitude of Resultant Vector
Since, R2 = ON 2 + CN 2
B
C
B
θ
O
R=
A+
B
β
θ
A
A
B cos θ
(a) Ax = 4 and Ay = 6 …(1)
A + B = Ax iˆ + Ay ˆj + Bx iˆ + By ˆj
ˆ
ˆ
⇒ A + B = ( Ax + Bx ) i + ( Ay + By ) j (
) (
)
So, Ax + Bx = 10 and Ay + By = 9 …(2)
From equation sets (1) and (2), we get
Bx = 6 and By = 3
So, B = 6iˆ + 3 ˆj
(b) B = 6 2 + 3 2 = 36 + 9 = 45 = 3 5 m
(c) Let B make an angle θ with x-axis, then from
the diagram
y
By
θ
0
⇒
tan θ =
Bx
By
Bx
=
2
⇒
R2 = ( OA + AN ) + CN 2 ⇒
R2 = A 2 + B2 + 2 AB cos θ R = R = A + B = A 2 + B2 + 2 AB cos θ
⇒
3 1
=
6 2
−1 ⎛ 1 ⎞
θ = tan ⎜ ⎟ , with x-axis
⎝ 2⎠
If two non zero vector are represented by the two
adjacent sides of a parallelogram then the resultant is given by the diagonal of the parallelogram
passing through the point of intersection of the two
vectors.
Direction of Resultant Vector
tan β =
CN
B sin θ
=
ON A + B cos θ
Conceptual Note(s)
Triangle Law and Parallelogram Law, both are equivalent. Since we know that a vector can be transported
parallel to its original direction, so their equivalency
holds good and is shown below.
θ
x
Parallelogram Law of Vector
Addition of Two Vectors
03_Vectors_Part 1.indd 7
B sin θ
B
Solution
3.7
B
A
TRIANGLE LAW
B
θ
A
PARALLELOGRAM LAW
Illustration 6
If A = 4iˆ − 3 ˆj and B = 6iˆ + 8 ˆj then find the magni tude and direction of A + B .
Solution
A + B = 4iˆ − 3 ˆj + 6iˆ + 8 ˆj = 10iˆ + 5 ˆj
| A + B|= (10)2 + (5)2 = 5 5
tan θ =
5 1
=
10 2
11/28/2019 6:56:29 PM
3.8 JEE Advanced Physics: Mechanics – I
Since it is given that ns is also a unit vector, therefore
we have
⎛ 1⎞
⇒ θ = tan −1 ⎜ ⎟
⎝ 2⎠
1 = 1 + 1 + 2 cos θ
5
A
⇒
+B
θ
cos θ = −
⇒ θ = 120° 1
2
The difference of the two vectors is given by
nd = nˆ 1 − nˆ 2
10
Illustration 7
In figure, a particle is moving in a circle of radius
r centred at O with constant speed v . What is the
change in velocity in moving from A to B ? Given
∠AOB = 60°.
⇒
nd2 = n12 + n22 − 2n1n2 cos θ = 1 + 1 − 2 cos ( 120° ) ⇒
⎛ 1⎞
nd2 = 2 − 2 ⎜ − ⎟ = 2 + 1 = 3
⎝ 2⎠
⇒
nd = 3 Illustration 9
Consider two vectors A and B inclined at an angle
θ . If A = B = R , then prove that
O
60°
VB
B
60°
A
VA
Solution
Change in velocity = Δv = vB − vA
vB − vA = v 2 + v 2 − 2v 2 cos ( 60° )
⇒
⎛ 1⎞
vB − vA = v 2 + v 2 − 2v 2 ⎜ ⎟ = v
⎝ 2⎠
Illustration 8
If the sum of two unit vectors is a unit vector, then
find the magnitude of difference of these two vectors.
Solution
Let n̂1 and n̂2 are the two unit vectors, then the sum
and difference of these two vectors be represented by
ns = nˆ 1 + nˆ 2 and nd = nˆ 1 − nˆ 2
⇒
ns2 = n12 + n22 + 2n1n2 cos θ = 1 + 1 + 2 cos θ
03_Vectors_Part 1.indd 8
⎛θ⎞
A + B = 2R cos ⎜ ⎟
⎝ 2⎠
(b)
⎛θ⎞
A − B = 2R sin ⎜ ⎟
⎝ 2⎠
Solution
Since vB − vA = vB2 + vA2 − 2vA vB cos θ
⇒
(a)
(a) Since A = B = 1
A + B = A 2 + B2 + 2 AB cos θ
⇒
A + B = R2 + R2 + 2R2 cos θ = 2R 1 + cos θ
⇒
2⎛θ⎞
Since 1 + cos θ = 2 cos ⎜⎝ 2 ⎟⎠ ⇒
⎛θ⎞
A + B = 2R cos ⎜ ⎟
⎝ 2⎠
(b) Similarly,
A − B = R2 + R2 − 2R2 cos θ = 2R 1 − cos θ
2⎛θ⎞
Since 1 − cos θ = 2 sin ⎜⎝ 2 ⎟⎠ ⇒
⎛θ⎞
A − B = 2R sin ⎜ ⎟
⎝ 2⎠
11/28/2019 6:56:37 PM
Chapter 3: Vectors
TRIANGLE INEQUALITY
Since a vector cannot have a resultant more than the
maximum value and less than the least value of the
resultant, so we have
Rmin ≤ R ≤ Rmax
⇒
A−B≤ R≤ A+B
If A > B , then A − B ≤ A + B ≤ A + B
A − B ≤ A+B ≤ A + B
A − B ≤ A+B & A+B ≤ A + B
{called Triangle Inequality}
POLYGON LAW OF VECTOR ADDITION
If a number of non zero vectors are represented by
the ( n − 1 ) sides of an n -sided polygon then the
resultant
D
E
B+
A
O
B
+B
B
R
A
Problem Solving Technique(s)
(a) If the vectors form a closed n sided polygon with all
the sides in the same order then the resultant is 0.
–C
B
A
C
A+
A+B+C+D
E
(c) Existence of Additive Identify
For every vector a , we have
a+0 = 0+a
i.e., Adding 0 i.e., null vector to any vector (say a )
does not changes the magnitude
of a as well as
its direction, because 0 has zero magnitude and
has an arbitrary/indeterminate direction.
(d) Existence of Additive Inverse
For a given vector a , there exists a vector ( − a )
such that
a + ( −a ) = 0
The vector ( − a ) is called the additive inverse of
a and vice versa.
C
C
C
D
A
O A + AB + BC + CD + D E = OE
Properties of Vector Addition
(a) Vector Addition is Commutative
For any two vectors a and b , we have
a+b = b+a
(b) Vector Addition is Associative
For any three vector a , b and c , we have
b + (c + a ) = c + (a + b ) = a + (b + c )
03_Vectors_Part 1.indd 9
B
A
Addition of Vectors
(b) To find the sum of any number of vectors we
must represent the vectors by the directed line
segments with the terminal point of the previous
vectors as the initial point of the next vector, then
the lines segment joining the initial point of the
first vector to the terminal
d
C
is given by the closing side or the nth side of the polygon taken in opposite order. So,
R = A+B+C+D+E
3.9
c
e
E
D
F
B
f
g
O
b
a
A
point of the last vector will represent the sum of
the vectors. Such that
(c) If the terminal of the last vector coincides with the
initial point of the first vector, then its sum will be
zero.
a+b+c +d+e + f +g=0
11/28/2019 6:56:42 PM
3.10 JEE Advanced Physics: Mechanics – I
MULTIPLICATION OF A VECTOR BY A
SCALAR
If m by any scalar and a be any vector then ma is
defined as a vector whose magnitude is m times
that of the magnitude of a and whose direction is
same as of a if m is positive and will be opposite to
that of a if m is negative.
Properties of Multiplication of
Vector by a Scalar
(a) If m = 0 (ZERO), then ma = 0 and not 0 .
(b) If m and n are two scalars and a be any vector
in space, then
m ( na ) = n ( ma ) = mn ( a )
(c) Vector multiplication is distributive over scalar
addition
a ( m + n ) = ma + na
where, m, n are any two scalars and a be any
vector in space.
(d) Scalar multiplication is distributive over vector
addition. Let a and b are any two vectors and
m be any scalar, then
m ( a + b ) = ma + mb
POSITION VECTOR
If a point O is fixed as the origin is space (or plane)
and P is any point, then OP is called the position
vector of P with respect to O . So, position vector of a
point P, from a convenient fixed origin O is a vector
drawn from O to P .
If it is said
that P is any point in space, then it
means that OP is a position vector or we can say that
position vector of point P is r (say) with respect to
some conveniently fixed origin O .
y
Conceptual Note(s)
A vector is an independent entity and can be transported any where keeping its Line of Action (LOA)
parallel to the same straight line, magnitude and sense
unchanged. i.e., A vector a will remain a till its magnitude and direction is retained.
Example: In many examples, when a force F is
applied on the block such that the block displaces
from its original position, the force F applied on block
remains same even after displacement w.r.t. the same
frame of reference.
Initial
position
F
A position vector is a fixed quantity with its initial
point as origin.
To Find AB If the Position Vectors of
Points A and B are Known
Let a and b are position vectors of points A and B
in plane (or space)
∴ OA = a and OB = b y
B
b
A
O
03_Vectors_Part 1.indd 10
x
Now, in ΔOAB , according to Triangular Law of
Vector addition, we have
OA + AB = OB
AB = OB – OA
1
x
(b – a)
a
r
O
F
BUT
⇒
P( r )
Final
position
⇒
2
⎛ Position Vector ⎞ ⎛ Position Vector ⎞
AB = ⎜
⎟⎠ − ⎜⎝
⎟⎠
of B
of A
⎝
11/28/2019 6:56:49 PM
Chapter 3: Vectors
SUBTRACTION OF VECTORS
Since, A − B = A + ( − B )
So, from above we conclude that subtraction
of vec
−
B OR
tors
is
actually
finding
the
resultant
of
A
and
B and − A .
Also, A + B = A 2 + B2 + 2 AB cos θ and
B sin θ
A + B cos θ
Since angle between A and B is θ , so angle between,
A and −B is ( 180 − θ ) .
tan b1 =
Rsum = A + B = d1
B
θ
180 –
–B
β1
β2
A
θ
Rdiff = A + (–B) = d2
⇒
A − B = A 2 + B2 + 2 AB cos ( 180 − θ ) Since, cos ( 180 − θ ) = − cos θ
⇒ A − B = A 2 + B2 − 2 AB cos θ So, d1 = A + B …(1)
d2 = A − B …(2)
Adding (1) & (2)
d1 + d2 1 A=
= d1 + d2
2
2
Subtracting (1) & (2)
d1 − d2 1 B=
= d1 − d2
2
2
(
)
(
)
RECTANGULAR COMPONENTS in
2-D SPACE
R
Consider
a
vector
inclined to the x-axis at angle θ .
If Rx and Ry are the components (projections) of R
along x-axis and y-axis then by parallelogram law
of ­vectors
R = Rx + Ry …(1)
Since Rx = Rx i and Ry = Ry j
⇒ R = Rx i + Ry j …(2)
î
î
î
Further
Rx = R cos θ …(3)
B sin θ
tan β1 =
A + B cos θ and tan β2 =
3.11
Ry = R sin θ …(4)
B sin ( 180 − θ )
A + B cos ( 180 − θ ) y
But sin ( 180 − θ ) = sin θ and cos ( 180 − θ ) = − cos θ
⇒
tan β2 =
B sin θ
A − B cos θ Ry 90 – θ
θ
x
Rx
Significance of Vector Addition and
Subtraction
If two vectors A and B are inclined to each other at
an angle q then
Rsum = A + B = d1 = Bigger Diagonal of parallelogram
Rdiff = A − B = d2 = Smaller Diagonal of parallelogram
Since, R = Rx i + Ry j
⇒ R = R cos θ i + sin θ j (
(
)
)
⇒
= R cos θ i + sin θ j RR
⇒
= cos θ i + sin θ j R
{∵ R = RRˆ }
{ R = 1}
Squaring (3) and (4) and adding, we get
Rx2 + Ry2 = R2 cos 2 θ + R2 sin 2 θ
03_Vectors_Part 1.indd 11
11/28/2019 6:56:56 PM
3.12 JEE Advanced Physics: Mechanics – I
⇒
Rx2 + Ry2 = R2 ⇒
R = Rx2 + Ry2
(
⇒
{using (3)}
⎛ Component ⎞
⎜ of vector ⎟
⎜
⎟
which ⎞
⎛ Angle
Rx ⎝ along x -axis ⎠
⎜
⎟
cos
R makes
=
=
R
⎜⎝ with x -axis ⎟⎠
⎛ Magnitude ⎞
⎜⎝ of vector ⎟⎠
Similarly sin θ =
Ry
cos ( 90 − θ ) =
R
Ry
{using (4)}
R ⎛ Component ⎞
⎜ of vector ⎟
⎜
⎟
which ⎞
⎛ Angle
Ry ⎝ along y -axis ⎠
⇒ cos ⎜ R makes ⎟ =
=
R
⎜⎝ with x -axis ⎟⎠
⎛ Magnitude ⎞
⎜⎝ of vector ⎟⎠
If R = A + B + C
Rx = Ax + Bx + Cx ⎫⎪
Then
in 2 D space
Ry = Ay + By + Cy ⎪⎭⎬
tan β =
Ry
=
Ay + By + Cy
Ax + Bx + Cx
where β is angle R makes with x-axis
Rx
Rx = Ax + Bx + Cx ⎫
Also, Ry = Ay + By + Cy ⎪⎬ 3D space
Rz = Az + Bz + Cz ⎪⎭
Problem Solving Technique(s)
(
(
x –
y +
x +
y +
x –
y –
x +
y –
r1st quad = r1 = r cosθi + sinθj
)
Illustration 10
Construct a vector of magnitude 6 unit making an
angle of 30° with y-axis.
Solution
30° with y-axis ⇒ 60° with x-axis.
Since, R = R cos θ i + sin θ j
where, θ is the angle which R makes with x-axis
⇒ R = 6 cos 60i + sin 60 j
⇒ R = 3 1i + 3 j ⇒ R = 3i + 3 3 j (
)
(
(
)
)
Illustration 11
A particle undergoes three successive displacements
in a plane. The first time it moves 4 m south-west the
second time 5 m east and the third time 6 m in direction
60° north of east. Draw a vector diagram and determine total displacement of particle from starting point.
Solution
iˆ OA = 4 cos ( 180° + 45° ) iˆ + 4 sin ( 180° + 45° ) ˆj
OA makes ( 180° + 45° ) angle with positive x-axis
and the same with positive y-axis so
x-component of OA = 4 cos ( 180° + 45° )
y-component of OA = 4 sin ( 180° + 45° )
A vector inclined at an angle q to x-axis in 1st and 4th
Quadrant OR at an angle q with -x axis in 2nd and 3rd
Quadrant is
03_Vectors_Part 1.indd 12
)
)
(
R
So, cos θ = x R
⇒
r2nd quad = r2 = r − cosθi + sinθj
r3rd quad = r3 = −r cosθi + sinθj
r4 th quad = r4 = r cosθi − sinθj
N
C
60°
O
W
6m
4m
A
)
E
B
5m
S
11/28/2019 6:57:03 PM
Chapter 3: Vectors
⇒ OA = −2 2iˆ − 2 2 ˆj
iˆ AB = 5 cos 0°iˆ = 5iˆ
iˆ BC = 6 cos 60°iˆ + 6 sin 60° ˆj = 3iˆ + 3 3 ˆj
⇒ OC = OA + AB + BC
⇒ OC = −2 2iˆ − 2 2 ˆj + 5iˆ + 3iˆ + 3 3 ˆj
⇒ OC = ( −2 2 + 5 + 3 ) iˆ + −2 2 + 3 3 ˆj
⇒ OC = +5 ⋅ 17 iˆ + 2.37 ˆj
(
)
(
(
)
tan α =
3.13
71
94
⇒ α = 37° So resultant is 118 N at 180 − 37 = 143°.
)
RECTANGULAR COMPONENTS in
3-D SPACE
Illustration 12
Four coplanar forces act on a body at point O as
shown in diagram by use of rectangular component
find direction and magnitude of resultant force.
In 3-D space, we have
R = Rx + Ry + Rz
R = Rx i + Ry j + Rz k
y
100 N
110 N
Ry
160 N
30°
20°
45°
O
80 N
x
x
Rz
z
The vectors and their components are as follows
cos α =
x component of
resultant vector
y component of
resultant vector
⇒
80
80
0
100
110 cos 45° = 71
100 sin 45° = 71
110
−110 cos 30° = −95
110 sin 30° = 55
160
−160 cos 20° = −150
−160 sin 20° = −55
Rx = 81 + 71 − 95 − 150 = −94 N
Ry = 0 + 71 + 55 − 55 = 71 N
Magnitude of resultant is
2
2
R = Rx2 + Ry2 = ( 94 ) + ( 71 ) = 118 N
03_Vectors_Part 1.indd 13
Rx
If R makes an angle a with x axis, b with y axis and
g with z axis, then
Solution
Magnitude
of
resultant
vector
R
Rx
R
cos α =
cos β =
Rx
=
R
Ry
⇒
cos β =
⇒
R
cos γ = z =
R
R
=
Ry
cos γ =
R
Rx
Rx2 + Ry2 + Rz2
Ry
Rx2 + Ry2 + Rz2
Rz
Rx2 + Ry2 + Rz2
Rz
R
=l
=m
=n
where l, m, n are called Direction Cosines of the
­vector R
2
2
2
cos α + cos β + cos γ =
Rx2 + Ry2 + Rz2
Rx2 + Ry2 + Rz2
=1
11/28/2019 6:57:09 PM
3.14 JEE Advanced Physics: Mechanics – I
Conceptual Note(s)
(a) If l, m, n are called Direction Cosines of the vector,
then l 2 + m2 + n2 = 1
cos2 α + cos2 β + cos2 γ = 1
(b) In 3-D space a vector of magnitude r making an
angle a with x-axis, b with y-axis and g with z-axis
can thus be written as
iˆ r = r ⎡⎣ ( cos α ) iˆ + ( cos β ) ˆj + ( cos γ ) kˆ ⎤⎦
illuStRAtion 13
A bird moves with velocity of 20 ms −1 in the direction making angle 60° with eastern line and 60°
with vertically upward. Represent the velocity vector
in rectangular form.
Solution
Velocity vector v makes angle α , β and γ with x-,
y- and z- axis respectively
∴
α = 60° and γ = 60°
Since, cos 2 α + cos 2 β + cos 2 γ = 1
⇒
cos 2 60° + cos 2 β + cos 2 60° = 1
⇒
⎛ 1⎞
⎛ 1⎞
2
⎜⎝ ⎟⎠ + cos β + ⎜⎝ ⎟⎠ = 1
2
2
⇒
cos 2 β = 1 −
⇒
cos β =
2
2
1
2
1
2
ˆ
Since
i v = v cos α iˆ + v cos β ˆj + v cos γ kˆ
⇒
⇒
1
1 ˆ
1
j + 20 × kˆ
v = 20 × iˆ + 20 ×
2
2
2
ˆ
ˆ
ˆ
v = 10i + 10 2 j + 10 k
Test Your Concepts-I
Based on Addition, Subtraction and Resolution
1. Can three vectors not in one plane give a zero
resultant? Can four vectors do?
2. The x and y components of vector A are 4 m and
6 m respectively. The x and y components of vector
A + B are 10 m and 9 m respectively. Calculate for
the vector B the following:
(a) its x and y components,
(b) its length and
(c) the angle its makes with x-axis.
3. If A = 4 iˆ − 3 ˆj and B = 6iˆ + 8 ˆj, obtain the sca lar magnitude and directions of A , B, ( A + B ),
( A − B ) and ( B − A ).
4. A particle has a displacement of 12 m towards
east and 5 m towards north and 6 m vertically
upwards. Find the magnitude of the sum of these
displacements.
5. Show that if two vectors are equal in magnitude,
their vector sum and difference are at right angles.
03_Vectors_Part 1.indd 14
(Solutions on page H.41)
6. Establish the following vector inequalities:
(a) a − b ≤ a + b
(b) a − b ≥ a − b
When does the equality sign apply?
7. Find the magnitude and direction of the resultant
of
following forces acting on a particle:
F1 = 3 2 Kgf due north-east, F2 = 6 2 Kgf due
south-east and F3 = 2 Kgf due north-west.
8. What does the statement a + b = a + b imply?
9. Two forces F1 and F2 acting at a point have a resultant R1. If F2 is doubled, the new resultant R2 is at
right angles to F1. Prove that R1 and F2 have the
same magnitude.
10. In figure, a particle is moving in a circle of radius
r centred at O with constant speed v. What is the
change in velocity in moving from A to B? Given
∠AOB = 40°.
11/28/2019 6:57:15 PM
Chapter 3: Vectors
(a)
⎛θ⎞
A + B = 2 cos ⎜ ⎟
⎝ 2⎠
(b)
⎛θ⎞
A − B = 2 sin ⎜ ⎟
⎝ 2⎠
O
40°
VB
B
40°
VA
A
11. Prove that the resultant of two vectors of equal
magnitude is equally inclined to either of the two
vectors.
12. Consider two unit vectors A and B inclined at an
angle q. Prove that
3.15
13. Can the magnitude of the resultant of two equal
vectors be equal to the magnitude of either of the
vectors? Explain your answer.
14. Prove that the vector r = iˆ + ˆj + kˆ is equally
inclined to all the three axis.
VECTOR MULTIPLICATION OF 2 VECTORS
MULTIPLICATION OF TWO VECTORS
INCLINED AT AN ANGLE θ
SCALAR PRODUCT
OR
DOT PRODUCT
VECTOR PRODUCT
OR
CROSS PRODUCT
A ⋅ B (read as A dot B)
A × B (read as A cross B)
A ⋅ B = ⎪A ⎢⎢B⎪ cos θ
A × B = ⎪A ⎢⎢B⎪ sin θ n
Where n is a unit vector giving the direction of A × B (or
obviously pointing towards A × B) and n is unit vector whose
direction can be found by RIGHT HAND THUMB RULE (RHTR)
according to which “Curl the fingers of RIGHT hand from 1st
vector to 2nd vector (i.e., from A to B in case of A × B and B
to A in case of B × A), then the direction of the thumb gives
direction of cross product or n”.
As a consequence of RHTR we conclude that it is a unit vector
normal to the plane containing (or having) A and B or we can
say that n is perpendicular to A and B simultaneously.
DOT PRODUCT
Geometrical Interpretation
Mathematically, dot product is defined as the product of the magnitudes of the vectors and cosine of the
angle between the vectors. So,
A ⋅ B = AB cos θ
Since A ⋅ B = A ( B cos θ )
⇒ A ⋅ B = A projection of B along A
03_Vectors_Part 1.indd 15
(
)
11/28/2019 6:57:17 PM
3.16 JEE Advanced Physics: Mechanics – I
B
A
OR
θ
A
B cos θ
⇒
⇒
⇒
θ
A cos θ
B
A⋅B
B cos θ =
A B ⋅ A = B ( A cos θ )
B ⋅ A = B × (Projection of A along B )
A⋅B
A cos θ =
B Physical Interpretation
Work done is dot product of force with displacement,
so we have
W = F ⋅ Δr
W = F ⋅ ( r2 − r1 ) where Δr = r2 − r1
1
r1
2
r2
(d) Dot product of vector with itself is equal to the
square of the magnitude of the vector
A ⋅ A = ( A )( A ) cos0
⇒ A ⋅ A = A2
(e) If θ = 180° i.e., Vectors are antiparallel, then
A ⋅ B = AB ( −1) { ∵ ( cos180° = −1) }
A ⋅ B = − AB
i.e., If two vectors are antiparallel then their dot
product equals the negative of simple product of
magnitudes of vectors.
(f) If θ = 90° i.e., vectors are perpendicular
A ⋅ B = AB ( 0 ) A ⋅ B = 0 i.e.,
Vectors perpendicular ⇔ Dot product = 0
(g) i ⋅i = i i cos0
k ⋅
k =1
Therefore, in general, i ⋅i = 1, j ⋅ j = 1, (h) i ⋅ j = i j cos90°
⇒ i ⋅ j = ( 1)( 1) ( 0 ) = 0
k = 0, k ⋅i = 0 or
Therefore, in general i ⋅ j = 0, j ⋅ j ⋅i = 0, k ⋅ j = 0, i ⋅ k=0
(i) If A = Axi + Ay j + Az k and B = Bxi + By j + Bz k
are any two vectors, then
A ⋅ B = Axi + Ay j + Az k ⋅ Bxi + By j + Bz k
k +
⇒ A ⋅ B = Axi ⋅ Bxi + By j + Bz (
(
Properties of Dot Product
(a) Dot product is commutative in nature
A⋅B = B⋅ A
(b) Dot product is distributive with respect to sum
A⋅(B ± C ) = A⋅B ± A⋅C
(c) If θ = 0° i.e. Vectors are parallel then
A ⋅ B = AB cos0 = AB i.e., when the vectors are
parallel A ⋅ B is just the simple product of the
magnitudes of A and B .
03_Vectors_Part 1.indd 16
(
)(
)
)
)
(
)
Ay j ⋅ Bxi + By j + Bz k + Az k ⋅ Bxi + By j + Bz k
⇒ A ⋅ B = Ax Bx + Ay By + Az Bz
(j) Since A ⋅ B = AB cosθ
and −1≤ cosθ ≤ 1
⇒ A ⋅ B is maximum at cosθ = 1( i.e., θ = 0° )
A ⋅ B is minimum at cosθ = −1( i.e., θ = 180° )
⇒ ( A ⋅ B )min ≤ A ⋅ B ≤ ( A ⋅ B )max
⇒ − AB ≤ A ⋅ B ≤ AB
(k) If A = Axi + Ay j + Az k and B = Bxi + By j + Bz k
and θ is ∠ between A & B.
A⋅B
Then cosθ =
where, A ⋅ B = AB cosθ
AB
11/28/2019 6:57:27 PM
Chapter 3: Vectors
2
A + B = ( A + B )⋅( A + B )
2
A − B = ( A − B )⋅( A − B )
(m) ( A + B ) ⋅ ( A − B ) = A ⋅ A − B ⋅ B + B ⋅ A − A ⋅ B
⇒ ( A + B ) ⋅ ( A − B ) = A2 − B
∵ A ⋅ A = A2 and B ⋅ B = B2
(n) Similarly
2
A + B + C = ( A + B + C )⋅( A + B + C )
2 ⇒ A + B + C = A⋅ A + A⋅B + A⋅C +
B ⋅ A + B ⋅B + B ⋅C + C ⋅ A + C ⋅B + C ⋅C
2
⇒ A + B + C = A2 + B 2 + C 2 +
2( A ⋅ B + B ⋅ C + C ⋅ A )
(o) If A = A i + A j + A k , has l , m , n as direction
(l)
{
z
y
z
}
1
1
⇒ cosθ = l1l2 + m1m2 + n1n2
Conceptual Note(s)
(a) Angle between two vectors A = Axi + Ay j + Az k
and B = B i + B j + B k is
x
cosθ =
y
z
Ax Bx + Ay By + Az Bz
A2x + A2y + Az2 B2x + B2y + Bz2
(b) If l1 m1 n1 and l2 m2 n2 are direction
cosines of A and B, then cosθ = l1l2 + m1m2 + n1n2 .
(c) If two vectors are perpendicular then
03_Vectors_Part 1.indd 17
l1l2 + m1m2 + n1n2 = 0
(d) Also
d ( ) dB dA ⎛ dA ⎞ ⎛ dB ⎞ A⋅B = A⋅
+B⋅
=⎜
⎟ ⋅ B + ⎜⎝
⎟⋅A
dt
dt ⎝ dt ⎠
dt ⎠
dt
Illustration 14
Three vectors A , B and C are such that A = B + C
and their magnitude are 5, 4, 3. Find angle between
A and C .
Solution
As A = B + C
⇒ B = A−C Now B ⋅ B = ( A − C ) ⋅ ( A − C )
B2 = A 2 + C 2 − 2 AC cos θ
1
cosines
If B = Bxi + By j + Bz k , has l2 , m2 , n2 as direction
cosines
Ax Bx + Ay By + Az Bz
Also, cosθ =
AB
A
AB
AB
y By
⇒ cosθ = x x +
+ z z
AB
AB
AB
⎛ A ⎞ ⎛ B ⎞ ⎛ Ay ⎞ ⎛ By ⎞ ⎛ Az ⎞ ⎛ Bz ⎞
⇒ cosθ = ⎜ x ⎟ ⎜ x ⎟ + ⎜
+
⎝ A ⎠ ⎝ B ⎠ ⎝ A ⎟⎠ ⎜⎝ B ⎟⎠ ⎜⎝ A ⎟⎠ ⎜⎝ B ⎟⎠
3.17
⇒
A 2 + C 2 − B2 52 + 3 2 − 4 2 18
=
=
= 0.6
2 AC
2×5×3
30
cos θ =
θ = cos −1 ( 0.6 )
Illustration 15
For what value of a are the vectors A = aiˆ − 2 ˆj + kˆ
and B = 2 a iˆ + a ˆj − 4 kˆ perpendicular to each other?
Solution
Now A and B are ⊥ to each other
⇒ A⋅B = 0 ⇒ A ⋅ B = a iˆ − 2 ˆj + kˆ ⋅ 2 a iˆ + a ˆj − 4 kˆ
2
⇒ A ⋅ B = 2a − 2a − 4 = 0 ∵ i ⋅ i = j ⋅ j = k ⋅ k = 1, i ⋅ j = j ⋅ k = k ⋅ i = 0
⇒ a2 − a − 2 = 0 (
)(
{
⇒
a2 − 2a + a − 2 = 0 ⇒
a( a − 2) + ( a − 2) = 0 ⇒
( a + 1)( a − 2 ) = 0 ⇒
a = −1 , a = 2 )
}
Illustration 16
Find the component of a = 2iˆ + 3 ˆj along the direction
of vector iˆ + ˆj .
11/28/2019 6:57:37 PM
3.18 JEE Advanced Physics: Mechanics – I
Solution
Given component of a in the direction of b is
ˆi r = ( a cos θ ) bˆ = ⎛⎜ a ⋅ b ⎞⎟ bˆ = ⎛ a ⋅ b ⎞ b
⎜⎝ 2 ⎟⎠
⎝ b ⎠
b
⇒
(
)(
)(
2iˆ + 3 ˆj ⋅ iˆ + ˆj ˆ ˆ
5
r=
i + j = iˆ + ˆj
2
2
( 1 )2 + ( 1 )2
)
(
)
5
So, component vector of a along b is r = iˆ + ˆj .
2
(
)
Test Your Concepts-II
Based on Dot Product
v x′ = 2 ms −1 and v y′ = 2 ms −1 . If both the balls start
moving from the same point, what is the angle
between their paths?
3. If A = 4iˆ + 6 ˆj − 3kˆ and B = −2iˆ − 5 ˆj + 7kˆ , find the
(a) direction cosines of A and B.
(b) angle between A and B .
4. Show
that
the
vectors
A = 3iˆ − 2 ˆj + kˆ ,
B = iˆ − 3 ˆj + 5kˆ and C = 2iˆ + ˆj − 4kˆ form a right
angle triangle.
(Solutions on page H.43)
5. Find the projection of A = 10iˆ + 8 ˆj − 6kˆ along
r = 5iˆ + 6 ˆj + 9kˆ .
6. Obtain the scalar product of the vectors ( 6, 2, 3 ),
( 2, − 9, 6 ) and also the angle between them.
7. Show that, for a vector u of constant magnitude,
du
= 0.
we have u ⋅
dt
8. Forces acting on a particle have magnitudes
5 N, 3 N and 1 N and act in the directions of the
vectors 6iˆ + 2 ˆj + 3kˆ , 3iˆ − 2 ˆj + 6kˆ , 2iˆ − 3 ˆj − 6kˆ
respectively. These remain constant while the particle is displaced from the point A ( 2, − 1, − 3 ) to
B ( 5, − 1, − 1) . Find the work done by the forces, the
unit of length being 1 m.
CRoSS PRoDuCt oR VECtoR PRoDuCt
y
Mathematically,
A × B = AB sin θ n
Direction of n is given by RHTR (stated earlier).
n̂ indicates direction of A × B and −n indicates
direction of B × A .
A × B = −B × A
So, cross product is Non-commutative in nature.
A × B = AB sin θ n
03_Vectors_Part 1.indd 18
n
A
θ
z
xy plane
1. Express the scalar product of two vectors in terms
of their rectangular components.
2. Two billiard balls are rolling on a flat table.
One has the velocity components v x = 1ms −1,
v y = 3 ms −1 and the other has components
B
e
xz
n
pla
x
–n
11/28/2019 6:57:43 PM
Chapter 3: Vectors
A × B = AB sin θ ⇒ B × A = BA sin θ ∴
A × B = B × A = AB sin θ A×B
Also, n̂ = , where, n indicates direction of A × B
A×B
3.19
⇒
θ
θ
A
Geometrical Interpretation of
Cross Product
Half of magnitude of cross product equals the area of
triangle with adjacent sides A and B .
1
1
Area = ( A ) ( B sin θ ) = ( AB sin θ )
2
2
1 ⇒ Area of triangle = A × B
2
⎛ perpendicular ⎞
Area of
⎛
⎞
⎜
⎟
(
)
⎜⎝ parallelogram ⎟⎠ = Base × ⎜ distance between ⎟
⎝ parallel sides ⎠
Area of
⎛
⎞
⎜⎝ parallelogram ⎟⎠ = A ( B siin θ ) = AB sin θ
1 Also, Area of parallelogram = A × B = d1 × d2
2
03_Vectors_Part 1.indd 19
A
Conceptual Note(s)
Conceptual Note(s)
(a) A × B = B × A = AB sinθ
(b) n is a new vector perpendicular to A as well as B.
n̂ ⋅ A = 0
{Perpendicular vectors have dot product = 0 }
n̂ ⋅ B = 0
Hence, ( A × B ) is a new vector perpendicular to
A as well as B
(c) n is perpendicular to A as well as B
(where n indicates direction of A × B )
So, A × B is a new vector ⊥ to A as well as B .
⇒ ( A × B )⋅ A = 0
⇒ (B × A )⋅ A = 0
⇒ ( A × B )⋅B = 0
⇒ (B × A )⋅B = 0
B
B sin θ
B sin θ
B
So, half of the modulus of cross product equals the
area of the triangle with adjacent sides A and B and
magnitude of cross product equals the area of the
parallelogram with adjacent sides A and B.
Physical Interpretation
(a) The torque due to a force F acting at a point with
position vector r about theorigin is τ = r × F .
The torque due to a force F acting at a point
with position vector r about another point hav
ing position vector r0 is τ = ( r − r0 ) × F .
τ (read as tau) i.e., Torque (in Physics)
r is the distance of point of application of force
from axis of rotation (A.O.R.) and F is force.
τ is also called the MOMENT OF FORCE.
(b) L = r × p
L = Angular momentum also called MOMENT
OF LINEAR MOMENTUM
p = Linear momentum = mv
p = mv
Since, Mass is a scalar therefore momentum can
be read as mass times velocity
iˆ
L = r × ( mv ) = m ( r × v ) = m x
vx
ˆj
y
vy
kˆ
z
vz
(c) v = ω × r , where v is the linear velocity of a par
ticle moving in a circle of radius vector r with
angular velocity ω .
(d) Magnetic force experienced by a charge q enter
ing a magnetic field with a v­elocity v is given by
Fmagnetic = q ( v × B )
11/28/2019 6:57:53 PM
3.20 JEE Advanced Physics: Mechanics – I
Properties of Vector Product/Cross Product
(a) Vector Product is Non commutative i.e.
A × B = −B × A
⇒ ( A × B ) + ( B × A ) = 0
i.e., cross product is position sensitive.
(b) Cross product is distributive with respect to sum
i.e.
A × (B + C ) = A × B + A × C
(c) If A and B are parallel i.e. θ = 0°, then
A × B = AB sin0
n
⇒ A × B = 0
Vectors parallel ⇔ cross-product equal to 0
(d) If A and B are antiparallel i.e. θ = 180°
A × B = AB sin( 180° )
n
{ ∵ sin180° = 0 }
⇒ A × B = 0 Vectors antiparallel ⇔ cross product equal to 0
(e) A × A = AA sin0
n
A× A=0
i.e., cross product of vector with itself is 0
(f) i ×i = 0, j × j = 0, k ×
k=0
i
Ay
⇒ A × B =i
By
k
–
j
For a Right handed triad system, curl fingers of
your right hand from
x to y, thumb gives direction of z
y to z, thumb gives direction of x
z to x, thumb gives direction of y
(h) If A = Axi + Ay j + Az k , B = Bxi + By j + Bz k , then,
k
Az
Bx
By
Bz
03_Vectors_Part 1.indd 20
2×2
Ax
Bx
Ax
Az
+
k
Bz 2 × 2
Bx
⇒ A × B = i ( Ay Bz − By Az ) − j ( Ax Bz − Az Bx ) +
Ay
By
2×2
k ( Ax B y − B x Ay )
Conceptual Note(s)
(a) A × B + C × A ≠ A × ( B + C )
Instead, A × B + C × A = A × B − A × C
⇒ A × B + C × A = A × ( B − C )
Please note that Cross Product is always position
sensitive. So be careful while changing the placement of vectors.
(b) Also we must know that
d ( ) dB dA dA dB
A×B =A×
+
×B=
×B + A×
dt
dt dt
dt
dt
Illustration 17
Find the area of triangle formed by tips of the vectors
a = iˆ − ˆj − 3 kˆ , b = 4iˆ − 3 ˆj + kˆ and c = 3iˆ − ˆj + 2kˆ .
Let ABC be the triangle formed by the tips of given
vectors. Then
BA = a − b = iˆ − ˆj − 3 kˆ − 4iˆ − 3 ˆj + kˆ = − 3iˆ + 2 ˆj − 4 kˆ
BC = c − b = 3iˆ − ˆj + 2kˆ − 4iˆ − 3 ˆj + kˆ = − iˆ + 2 ˆj + kˆ
(
(
(g) i × j = k, k ×i = j, j × k = i
j
Ay
Bz
−j
Solution
+
i
A × B = Ax
Az
⇒
⇒
⇒
⇒
⇒
⇒
3× 3
⇒
) (
) (
)
)
iˆ ˆj kˆ
BA × BC = −3 2 −4
−1 2 1
BA × BC = iˆ ( 2 + 8 ) − ˆj ( −3 − 4 ) + kˆ ( −6 + 2 )
BA × BC = 10iˆ + 7 ˆj − 4 kˆ
2
2
2
BA × BC = ( 10 ) + ( 7 ) + ( −4 ) BA × BC = 165 = 12.8 1 Area of ΔABC = BA × BC
2
1
Area of ΔABC = × 12.8 = 6.4 sq. unit
2
11/28/2019 6:58:01 PM
Chapter 3: Vectors
Illustration 18
If a particle of mass m is moving with constant velocity v parallel to x-axis in x -y plane as shown in
­figure, calculate its angular momentum with respect
to origin at any time t .
m
b
v
r
We know that
Angular momentum L = r × p
OR
kˆ
z
pz
As motion is in x -y plane
L = kˆ ( xpy − ypx )
Whether the vectors A and B are parallel or anti­
parallel then in both the cases
A×B = 0
HOW TO REMOVE THE CONFUSION?
If A = kB ( k > 0 ), then vectors are parallel and if
A = −kB ( k > 0 ), then vectors are antiparallel.
Solution
ˆj
y
py
3 3 9
= = =k
1 a 3
9a = 9 a = 1
⇒
CONFUSION?
x
iˆ
L= x
px
Ax Ay Az
=
=
=k
Bx
By
Bz
⇒
⇒
y
3.21
Let A = Axi + Ay j + Az k
B = Bxi + By j + Bz k
{∴ z = 0 and pz = 0 }
Here x = vt
y = b px = mv py = 0
L = kˆ [ vt × 0 − bmv ] = − ( mbv ) kˆ
If
Ax Ay Az
=
=
= +k ( k > 0 ), then A parallel to B,
B x B y Bz
A x A y Az
=
=
= −k ( k > 0 ), then A anti­
B x B y Bz
parallel to B.
else if
OR
A⋅B
Find cosθ i.e. angle between two vectors cosθ =
AB
If cosθ = 1, then vectors are parallel
If cosθ = −1, then vectors are antiparallel
From Result we can conclude that, if motion is in
x-y plane angular momentum is always directed along
z-axis i.e., angular momentum is always perpendicular
to plane of motion.
Directions
Illustration 19
The example below indicates the method to read and
express directions.
Find the value of a for which the vectors 3iˆ + 3 ˆj + 9kˆ
and iˆ + ajˆ + 3 kˆ are parallel.
N
j
E
W N
SW
03_Vectors_Part 1.indd 21
N N
E
i
SE
Let A = 3iˆ + 3 ˆj + 9kˆ
B = iˆ + ajˆ + 3 kˆ
Now A B so we have
W
Solution
S
11/28/2019 6:58:08 PM
3.22 JEE Advanced Physics: Mechanics – I
N
NNW
NNE
NNE: North of North East.
NE
NW
ENE: East of North East.
ENE
WNW
W
E
WSW
ESE
SSW: South of South West.
WSW: West of South West.
180 – α
NNW: North of North West.
SSE
SSW
SSE: South of South East.
WNW: West of North West.
SE
SW
ESE: East of South East.
Pre-multiplying both sides by a
a × ( a + b ) = −a × c
⇒ 0 + a × b = −a × c ⇒ a × b = c × a …(3)
α
S
(a) NW OR SW OR SE OR NE means 45° with
either of the axis.
(b) 30°NW means 30° towards the north of west
(c) 35°SW means 35° towards the south of west.
LAMI’S THEOREM
Statement
In any ΔABC with sides a , b , c , if α , β and γ be
the respective angles containing the sides, then
β
γ
180 – β
180 – γ
a
Pre-multiplying both sides of (2) by b
b×(a+ b) = b×c
⇒ b × a + b × b = −b × c ⇒ − a × b = −b × c ⇒ a × b = b × c …(4)
Taking magnitude, we get
a×b = b×c = c×a
A
γ
β
α
B
OR
For any triangle the ratio of the sine of the angle containing the side to the length of the side is a constant.
Proof
For a triangle whose three sides are in the same order
we establish the Lami’s Theorem in the following
manner. For the triangle shown all three sides are in
same order, so
a + b + c = 0 …(1)
⇒ a + b = − c …(2)
03_Vectors_Part 1.indd 22
b
From (3) and (4), we get
a×b = b×c = c×a
sin α sin β sin γ
=
=
a
b
c
C
c
⇒
ab sin ( 180 − γ ) = bc sin ( 180 − α ) = ca sin ( 180 − β ) ⇒
ab sin γ = bc sin α = ca sin β Dividing throughout by abc , we have
⇒
sin α sin β sin γ
=
=
a
b
c SCALAR TRIPLE PRODUCT (STP)
Let a , b and c be three vectors, then the scalar triple
product can be written as
( a × b ) ⋅ c i.e., ( a CROSS b ) DOT c
It can also be written as [ a b c ]
11/28/2019 6:58:15 PM
Chapter 3: Vectors
Conceptual Note(s)
Scalar Triple Product (STP) is a scalar product and thus
has no direction.
GEOMETRICAL INTERPRETATION OF
SCALAR TRIPLE PRODUCT
Geometrically, the scalar triple product [ a b c ]
represents the volume of ­
parallelopiped whose
­coterminus edges a , b and c form a right handed
system of vectors.
b
(a) Cyclic Permutation of a , b and c does not
change the value of the scalar triple product
⎡ a b c ⎤⎦ = ⎣⎡ b c a ⎤⎦ = ⎣⎡ c a b ⎤⎦
⎣
(b) An antisymmetric or acyclic permutation
changes sign only and not magnitude
⎡⎣ a b c ⎤⎦ = − ⎡⎣ a c b ⎤⎦ = − ⎡⎣ b a c ⎤⎦
Conceptual Note(s)
The position of dot and cross can be interchanged
keeping the cyclic order same. With such combinations, 12 ­different combinations are possible.
(c) Scalar Triple Product in Components Form
If a , b and c are any three vectors, then,
a = ax i + ay j + az k ,
b = bx i + by j + bz k
and
c = cx i + cy j + cz k
03_Vectors_Part 1.indd 23
( a × b ) ⋅ c = bx
ay
az
by
bz
cx
cy
cz
Illustration 20
( 4i + 5j + k ) , − ( j + k ) ,
Prove that the four points
Solution
Properties of Scalar Triple Product
⇒
(d) For any three vectors a , b and c and scalar l,
we have
⎡⎣ λ a b c ⎤⎦ = ⎣⎡ a λ b c ⎤⎦ =
⎡⎣ a b λ c ⎤⎦ = λ ⎣⎡ a b c ⎤⎦
(e) If ⎡⎣ a b c ⎤⎦ = 0, then atleast two of the three vectors are collinear, equal or parallel.
(f) If ⎡⎣ a b c ⎤⎦ = 0, then the vectors a , b and c are
coplanar.
( 3i + 9j + 4k ) and 4 ( −i + j + k ) are coplanar.
c
a
3.23
ax
For four points (i.e., three vectors) to be collinear, let
us first find three vectors by t­aking one point as the
origin. So, let the origin be at the first vector, then
A = − ˆj − kˆ − 4iˆ + 5 ˆj + kˆ = −4iˆ − 6 ˆj − 2kˆ
B = 3iˆ + 9 ˆj + 4 kˆ − 4iˆ + 5 ˆj + kˆ = −iˆ + 4 ˆj + 3 kˆ
(
(
and
) (
)
) (
)
C = −4iˆ + 4 ˆj + 4 kˆ − 4iˆ + 5 ˆj + kˆ = −8iˆ − ˆj + 3 kˆ
For these to be coplanar, we must have A ⋅ ( B × C ) = 0
(
) (
)
−4 −6 −2
)
(
3
Now, A ⋅ B × C = −1 4
−8 −1 3
⇒ A ⋅ ( B × C ) = −4 ( 12 + 3 ) + 6 ( −3 + 24 ) − 2 ( 1 + 32 )
⇒ A ⋅ ( B × C ) = −60 + 126 − 66
⇒ A⋅(B × C ) = 0
Hence, we must say that the points or the vectors are
coplanar.
VECTOR TRIPLE PRODUCT (VTP)
Let a , b and c be three vectors, then the vector triple product is
( a × b ) × c or a × ( b × c )
11/28/2019 6:58:22 PM
3.24 JEE Advanced Physics: Mechanics – I
and is defined as
( a × b ) × c = ( a ⋅ c ) b − ( b ⋅ c ) a
OR a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
Problems Solving technique(s)
If a × ( b × c ) can also be written as × ( × ),
then it is defined as
× ( × ) = ( ⋅ ) − ( ⋅ )
i.e., a × ( b × c ) = ( a ⋅ c ) b − ( a ⋅ b ) c
Read as “ a cross b cross c equals a dot c times b
minus a dot b times c”.
Conceptual Note(s)
If r = a × ( b × c ), then r is a vector perpendicular to
a and lies in the plane of b × c .
Test Your Concepts-III
Based on Cross Product, Scalar and Vector triple Product
1. If a force F = 3iˆ + 5 ˆj − 2 kˆ N acts at a point
defined by 7 iˆ − 2 ˆj + 5 kˆ m, find the torque:
(
2.
3.
4.
5.
6.
7.
(
)
)
(a) about the origin, and
(b) about the point ( 0, − 1, 0 ) .
Express the vector product of two vectors in terms
of their rectangular components.
2
Prove that a × b = a2 b2 − ( a ⋅ b )
Show that ( a + b ) × ( a − b ) = −2 ( a × b ) and use
this result to find the area of a parallelogram whose
diagonals are iˆ − 2 ˆj − 3kˆ and 2iˆ − 3 ˆj + 2kˆ .
If L and L ′ , are two length vectors, what physical
quantity does [ L × L ′ ] represent?
If A × B = 0 and A ⋅ B = 0 , does it imply that one of
the vectors A or B must be a null vector?
The particle of mass m is projected at t = 0 from
a point O on the ground with speed v0 at an
angle 45° to the horizontal as shown in figure.
Compute the magnitude and direction of the
angular momentum of the particle about the point
03_Vectors_Part 1.indd 24
(Solutions on page H.44)
⎛ 0.2v02 ⎞
ˆj when the
+
i
g ⎟⎠ ⎜⎝ g ⎟⎠
velocity of the particle is
iˆ v = ( 0.7v ) iˆ − ( 0.3v ) ˆj.
⎛ 0.7v02 ⎞
ˆ
O at position
iˆ r = ⎜
⎝
0
0
y
v0
45°
O
x
8. The force on a positively charged particle is given
by F = qE + qv × B . In a certain space there is a
magnetic field B along z-axis and an electric field
along y-axis. A positively charged particle is projected into this space. Find the direction and magnitude of minimum velocity so that it may pass on
undeviated.
9. Find the moment of force F = iˆ + ˆj + kˆ acting at
point ( −2, 3, 4 ) about the point ( 1, 2, 3 ) .
10. Prove that a × ( b × c ) + b × ( c × a ) + c × ( a × b ) = 0.
11/28/2019 6:58:30 PM
Chapter 3: Vectors
3.25
Solved Problems
Problem 1
N
Let F1 and F2 be the two forces where F1 < F2
Here, F1 + F2 = 18 …(1)
The statement to the question can be diagrammatically drawn as shown in Figure.
F2
12
F1
So, the diagram clearly shows that 12 is the resultant
of F1 and F2 (taken in same order). Also we observe
that 12 i.e., resultant is perpendicular to F1 . So, from
the diagram, by using Pythagoras ­theorem, we get
W
A
O
⇒
( F2 − F1 ) ( F2 + F1 ) = 144 ⇒
( F2 − F1 ) ( 18 ) = 144
⇒
F2 − F1 = 8 …(2)
F1 = 5 and F2 = 13
Problem 2
E
i
S
Solution
N
Change in velocity = Δv = v f − vi
v1
v
(a) Now, for half revolution.
If v3 = v and v1 = −v
as their magnitudes are equal but
directions opposite.
⇒ Δv = v3 − v1 = v − ( −v ) = 2v
Δv = 2v directed south.
(b) For, quarter revolution
Δv = v2 − v1 and θ = 90°
v
v3
S
Δv
N
v1
v
v2
v
ϕ
Δv
E
v
–v1
S
So, from (1) and (2), we get
E
v3
W
F22 − F12 = 144 N N
W N
C
F22 = 144 + F12
⇒
j
v1
SE
Solution
v2
SW
The sum of the magnitudes of two forces acting at
a point is 18 and the magnitude of their resultant is
12. If the resultant is at 90° with the force of smaller
magnitude, what are the magnitudes of forces?
B
Δv = v 2 + v 2 + 2vv cos 90° = 2v
⎛ v⎞
and ϕ = tan −1 ⎜ ⎟ = 45°
⎝ v⎠
So, Δv = ( 2 ) v south-west
Problem 3
A body is moving with uniform speed v in a horizontal circle in anticlockwise direction starting from A as
shown in figure. Calculate the change in velocity in
If vectors A , B and C have magnitudes 8, 15 and 17
unit and A + B = C , find the angle between A and B .
(a) half revolution
(b) first quarter revolution.
Solution
03_Vectors_Part 1.indd 25
A+B = C
11/28/2019 6:58:36 PM
3.26 JEE Advanced Physics: Mechanics – I
Taking magnitude both sides
A+B = C
{Magnitudes of equal vectors is also equal}
Squaring both sides, we get
2 2
A+B = C
2
⇒ ( A + B )⋅( A + B ) = C ⋅C ∵ A⋅ A = A
{
⇒
A 2 + B2 + 2 AB cos θ = C 2
⇒
cos θ =
}
C 2 − A 2 − B2
2 AB
17 2 − { ( 15 ) + ( 8 ) }
2 × 15 × 8
⇒ cos θ = 0
⇒ θ = 90°
⇒
2
cos θ =
2
(
(
Problem 4
If the unit vectors a and b are inclined at angle θ
⎛θ⎞
prove that a − b = 2 sin ⎜ ⎟ .
⎝ 2⎠
Since we know that,
2
2
2
aˆ − bˆ = aˆ + bˆ − 2 aˆ bˆ cos θ
{∵ aˆ = bˆ = 1}
a − b
2
⇒
a − b
2
⇒
2
⎛θ⎞
⎛θ⎞
a − b = 2( 1− cos θ )= 2 × 2 sin 2 ⎜ ⎟ = 4 sin 2 ⎜ ⎟
⎝ 2⎠
⎝ 2⎠
= 1 − cos θ − cos θ + 1 )
(
Hence, vector area of triangle =
(
) (
(
)
1
ˆ
ˆ
ˆ
⇒ Area = 2 10i + 7 j − 4 k
)
1 )
( BA × BC
2
)
Magnitude of area of the triangle is
= 2 − 2 cos θ
Taking square root both sides.
Area =
1
1
100 + 49 + 16 =
165 square units
2
2
Problem 6
A particle is in equilibrium under the action of three
forces. Prove that each force bears a constant ratio
with the sine of angle between the other two.
⎛θ⎞
a − b = 2 sin ⎜ ⎟
⎝ 2⎠
Solution
Problem 5
(a) Find the area of the parallelogram determined by
the vectors a = 3iˆ + 2 ˆj and b = 2 ˆj − 4 kˆ .
(b) Find the area of the triangle whose vertices are
( 1, − 1, − 3 ) , ( 4, − 3, 1 ) and ( 3, − 1, 2 ) .
Let P , Q and R be the three forces acting at a point
O . Since particle is in equilibrium.
P
Q
γ
β
Solution
(a) Vector area
of parallelogram with adjacent sides
a and b is
a × b = 3iˆ + 2 ˆj × 2 ˆj − 4 kˆ
(
03_Vectors_Part 1.indd 26
)
1
ˆ
ˆ
ˆ
ˆ
ˆ ˆ
⇒ Area = 2 −3i + 2 j − 4 k × −i + 2 j + k
Solution
⇒
iˆ ˆj kˆ
⇒ a × b = 3 2 0
0 2 −4
ˆ
ˆ
ˆ
⇒ a × b = i [ −8 − 0 ] − j [ −12 − 0 ] + k [ 6 − 0 ]
ˆ
ˆ
ˆ
⇒ a × b = −8i + 12 j + 6 k
∴ Magnitude of the area of parallelogram
a × b = −8iˆ + 12 ˆj + 6 kˆ
⇒ a × b = 64 + 144 + 36 = 2 61
(b) Position vectors of the vertices A, B and C of
the triangle ABC are a = iˆ − ˆj − 3 kˆ ,
b = 4iˆ − 3 ˆj + kˆ and c = 3iˆ − ˆj + 2kˆ
BA
= a − b = −3iˆ + 2 ˆj − 4 kˆ
and BC = c − b = −iˆ + 2 ˆj + kˆ
) (
)
α
R
11/28/2019 6:58:47 PM
Chapter 3: Vectors
P + Q + R = 0 …(1)
⇒ ( P + Q ) = − R …(2)
Taking cross product, with P on both sides of (2)
P × ( P + Q ) = −P × R
⇒ P × P + P ×Q = R× P
⇒ P × Q = R × P …(3)
Similarly taking cross product with Q on both sides
of (3), we get
P × Q = Q × R …(4)
From (3) and (4)
P ×Q = Q× R = R× P
⇒
PQ sin γ QR sin α RP sin β
=
=
PQR
PQR
PQR
⇒
θ
⎛ β⎞
tan ⎜ ⎟ =
⎝ 2⎠
Solution
Let the two vectors be A and B , inclined at an angle
θ . Also we have been given that A = B . If β be
the angle which the resultant (of A and B ) makes
with A , then
tan β =
⇒
B sin θ
B + B cos θ
sin θ
tan β =
…(1)
1 + cos θ
03_Vectors_Part 1.indd 27
θ
β /2
A
B
sin θ
2
B
A + cos θ
2
Since A = B , so
sin θ
⎛ β⎞
…(2)
tan ⎜ ⎟ =
⎝ 2 ⎠ 2 + cos θ
Now, from concepts of trigonometric functions, we
have
⎛ β⎞
2 tan ⎜ ⎟
⎝ 2⎠
tan β =
⎛ β⎞
1 − tan 2 ⎜ ⎟
⎝ 2⎠
⇒
Problem 7
Two vectors of equal magnitude are inclined at an
angle θ . When one of them is halved, the angle
which the resultant makes with the other is also
halved. Find θ .
β
When B is halved, then still the angle between A
B
and
is θ . However, the new resultant is now
2
β
inclined to A at an angle . So, we get
2
P
Q
R
=
=
sin α sin β sin γ
The Lami’s Theorem is very useful in the chapters to
come. So please understand the theorem carefully so
that you can apply it at the place required.
B
2
A
PQ sin γ = QR sin α = RP sin β
Dividing both sides by PQR , we get
R
B
3.27
sin θ
=
1 + cos θ
⎛ sin θ ⎞
2⎜
⎝ 2 + cos θ ⎟⎠
1−
sin 2 θ
( 2 + cos θ )2
⇒
2 ( sin θ ) ( 2 + cos θ )
sin θ
=
1 + cos θ 4 + cos 2 θ + 4 cos θ − sin 2 θ
⇒
4 + cos 2 θ + 4 cos θ − sin 2 θ =
( 4 + 2 cos θ ) ( 1 + cos θ )
⇒
4 + cos 2 θ + 4 cos θ − sin 2 θ =
4 + 2 cos 2 θ + 6 cos θ − cos 2 θ − 2 cos θ − sin 2 θ = 0
⇒
−2 cos θ − ( sin 2 θ + cos 2 θ ) = 0
⇒
−2 cos θ − 1 = 0 ⇒
cos θ = −
{∵ sin 2 θ + cos2 θ = 1}
1
2
⇒ θ = 120°
11/28/2019 6:58:57 PM
3.28 JEE Advanced Physics: Mechanics – I
Problem 8
b
b
1
cos θ 2 = =
=
3b
3
a+b+c
If a , b and c are three mutually perpendicular
­vectors of equal magnitude, prove that a + b + c is
equally inclined to a , b and c .
Solution
Given that a = b = c and the three vectors are
mutually perpendicular. So,
a⋅b = b ⋅c = c ⋅a = 0
Let θ1 be the angle between a and a + b + c . Then
a ⋅ ( a + b + c ) = a a + b + c cos θ1
⇒ a ⋅ a + a ⋅ b + a ⋅ c = a a + b + c cos θ1
⇒ a 2 = a a + b + c cos θ1
a
cos θ1 = …(1)
a+b+c
Similarly, if θ 2 be the angle between b and a + b + c
and θ 3 is the angle between c and a + b + c , then
⇒
b
cos θ 2 = …(2)
a+b+c
c
cos θ 3 = …(3)
a+b+c
Now, let us find the value of a + b + c . Since we
know that dot product of the vector with itself is
equal to the square of its magnitude, so
( a + b + c ) ⋅ ( a + b + c ) = a + b + c 2
2
⇒ a ⋅ a + b ⋅b + c ⋅c + 2( a ⋅b + b ⋅c + c ⋅ a ) = a + b + c
Since a ⋅ b = b ⋅ c = c ⋅ a = 0 , so we get
2
a2 + b 2 + c2 = a + b + c
2
Again a = b = c , so a + b + c = 3 a 2
⇒ a + b + c = 3 a …(4)
From (1), (2), (3) and (4), we get
a
a
1
cos θ1 = =
=
3a
3
a+b+c
03_Vectors_Part 1.indd 28
c
c
1
cos θ 3 = =
=
3c
3
a+b+c
Since cos θ1 = cos θ 2 = cos θ 3 =
1
3
⎛ 1 ⎞
⇒ θ1 = θ 2 = θ 3 = cos −1 ⎜
⎝ 3 ⎟⎠
Hence the vector a + b + c is equally inclined to a , b
and c .
Problem 9
For a vector a in 3-D space, find the value of the
2
2
2
expression iˆ × a + ˆj × a + kˆ × a
Solution
Let the vector a make angle α , β and γ with x , y
and z axis respectively. Then
2
iˆ × a = iˆ a sin 2 α
⇒
2
iˆ × a = a 2 sin 2 α
Similarly
ˆj × a 2 = a 2 sin 2 β and kˆ × a 2 = a 2 sin 2 γ
2
2
2
Now iˆ × a + ˆj × a + kˆ × a =
a 2 ( sin 2 α + sin 2 β + sin 2 γ ) …(1)
Now since
cos 2 α + cos 2 β + cos 2 γ = 1
⇒
( 1 − sin 2 α ) + ( 1 − sin 2 β ) + ( 1 − sin 2 γ ) = 1
⇒
sin 2 α + sin 2 β + sin 2 γ = 2 …(2)
From (1) and (2) we get
2
2
2
iˆ × a + ˆj × a + kˆ × a = 2 a2
Problem 10
Let a, b and c be three vectors such that a + b + c = 0.
If a = 3, b = 4 and c = 5, then find the value of
( a ⋅ b + b ⋅ c + c ⋅ a ).
11/28/2019 6:59:08 PM
Chapter 3: Vectors
Solution
Since a + b + c = 0
⇒ a + b = −c
⇒ a ⋅ c + b ⋅ c = −c ⋅ c
⇒ a ⋅ c + b ⋅ c = − c 2 …(1)
Similarly a + c = −b
⇒ a ⋅ b + c ⋅ b = −b 2 …(2)
03_Vectors_Part 1.indd 29
3.29
Again b + c = − a
⇒ b ⋅ a + c ⋅ a = − a 2 …(3)
Adding (1), (2) and (3), we get
2 ( a ⋅ b + b ⋅ c + c ⋅ a ) = − ( a2 + b 2 + c2 )
⇒
⇒
( a ⋅ b + b ⋅ c + c ⋅ a ) = − 1 ( 32 + 42 + 52 ) = −25
2
( a ⋅ b + b ⋅ c + c ⋅ a ) = −25
11/28/2019 6:59:10 PM
3.30 JEE Advanced Physics: Mechanics – I
Practice Exercises
Single Correct Choice Type Questions
This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
A particle starts from rest at the origin with a con
stant acceleration a = 2iˆ + 8 ˆj − 6 kˆ ms −2 . Its position at
7.
The sum, difference and cross product of two vectors
A and B are mutually perpendicular if
(A) A and B are perpendicular to each other
(B)
A and B are perpendicular but their magnitudes
are arbitrary
(C) A = B and their directions are arbitrary
(D) A ⊥ B and A = B
8.
Three forces 9, 12 and 15 N acting at a point are in equilibrium. The angle between 9 N and 15 N is
t = 5 s is
( 25iˆ + 100 ˆj − 75kˆ ) m (B) ( 25iˆ − 100 ˆj − 75kˆ ) m
(C) ( 100iˆ − 25 ˆj + 75kˆ ) m (D) ( 25iˆ − 100 ˆj + 75kˆ ) m
Three forces F1 = ( 3iˆ + 2 ˆj − kˆ ) N, F 2 = ( 3iˆ + 4 ˆj − 5kˆ ) N
and F 3 = A ( iˆ + ˆj − kˆ ) N act simultaneously on a par(A)
2.
ticle. In order that the particle remains in equilibrium,
the value of A should be
(B) 6
(A) −6
(C) 9
(D) −9
3.
The magnitude of vector product of two non zero vec
tors A and B is zero. The scalar product of A and
A + B is equal to
(A) zero
(B) AB
(C) A 2
4.
5.
6.
(D) A 2 + AB
The magnitude of resultant of two equal forces is equal
to either of two forces. The angle between forces is
(A)
π
3
(B)
2π
3
(C)
π
2
(D)
3π
4
The condition under which the vectors ( a + b ) and
( a − b ) are parallel is
(A) a ⊥ b
(B) a = b
(C) a ≠ b
(D) a b
The two vectors A and B that are parallel to each
other are
(A) A = 3iˆ + 6 ˆj + 9kˆ , B = iˆ + 2 ˆj + 3 kˆ
(B) A = 3iˆ − 6 ˆj + 9kˆ , B = iˆ + 2 ˆj + 3 kˆ
(C) A = iˆ + 2 ˆj + 3 kˆ , B = iˆ + 2 ˆj − 3 kˆ
(D) A = 3iˆ + 6 ˆj − 9kˆ , B = iˆ − 2 ˆj − 3 kˆ
03_Vectors_Part 2.indd 30
9.
⎛ 3⎞
(A) cos −1 ⎜ ⎟
⎝ 5⎠
(B)
⎛ 4⎞
cos −1 ⎜ ⎟
⎝ 5⎠
⎛ 3⎞
(C) π − cos −1 ⎜ ⎟
⎝ 5⎠
⎛ 4⎞
(D) π − cos −1 ⎜ ⎟
⎝ 5⎠
2
The condition that ( a ⋅ b ) = a 2b 2 is satisfied when
(A) a b
(B) a ≠ b
(C) a ⋅ b = 1
(D) a ⊥ b
10. The sides of the triangle representing three force vectors are in the ratio 2000 : 1732 : 1000. The angles of the
triangle (in degrees) are
(A) 70, 65, 45
(B) 80, 55, 45
(C) 90, 60, 30
(D) 90, 61, 39
11. A force F = −3iˆ + ˆj + 5kˆ N acts at a point ( 7 , 3 , 1 ) .
The torque about the origin (0, 0, 0) will be
(A) 14iˆ − 38 ˆj + 16 kˆ
(B) 14iˆ + 38 ˆj − 16 kˆ
(
)
(C) 14iˆ − 38 ˆj − 16 kˆ
(D) 14iˆ + 38 ˆj + 16 kˆ
12. A vector C = A − B has a magnitude equal to A + B ,
the angle between A and B is
π
(A) zero
(B)
2
(C) π
(D) 2π
13. The vector having module equal to 3 and perpendicu
lar to two vectors A = 2iˆ + 2 ˆj + kˆ and B = 2iˆ − 2 ˆj + 3 kˆ is
(
)
(C) − ( 3iˆ − ˆj − 3 kˆ )
(A) ± 2iˆ − ˆj − 2kˆ
(
(B)
± 3iˆ + ˆj − 2kˆ
(D)
( 3iˆ − ˆj − 3kˆ )
)
11/28/2019 6:56:15 PM
Chapter 3: Vectors
14. The vector sum of three vectors a , b and c is zero. If
î and ĵ are unit vectors in the directions of a and b
respectively, then
(A) c is in the plane of î and ˆj .
(
)
c is along iˆ × ˆj .
(C) c is along iˆ.
(D) c is along ˆj .
(B)
15. 12 coplanar forces (all of equal magnitude) maintain a
body in equilibrium, then the angle between any two
adjacent forces is
(A) 15°
(B) 30°
(C) 45°
(D) 60°
16. If F1 × F2 = F1 ⋅ F2 , then F1 + F2 is
(A) F1 + F2
(B)
2 2
F1 + F2
(C)
2 2 F1 F2
F1 + F2 +
2
(D)
2 2
F1 + F2 + 2 F1 F2
17. The value of p so that the vectors 2iˆ − ˆj + kˆ , iˆ + 2 ˆj − 3 kˆ
and 3iˆ + pjˆ + 5kˆ are coplanar should be
(A) 16
(B)
−4
(D) −8
18. ABCD is a quadrilateral. Forces BA , BC , CD and DA
act at the point. Their resultant is
(A) 2AB
(B) 2DA
(C) 2BC
(D) 2BA
(C) 4
19. Which of the following group of forces will not accelerate a body?
(A) 5 N, 10 N, 12 N
(B) 5 N, 10 N, 16 N
(C) 8 N, 10 N, 20 N
(D) 7 N, 5 N, 15 N
20. Consider the two vectors A and B . The magnitude of
their sum, i.e. A + B , if A > B
(A) is equal to A + B
(B) cannot be less than A + B
(C) cannot be greater than A + B
(D) must be equal to A − B
03_Vectors_Part 2.indd 31
3.31
21. Three vectors P , Q and R are such that P = Q ,
R = 2 P and P + Q + R = 0 . The angles between
P and Q , Q and R and P and R are respectively (in
degrees)
(A) 90, 135, 135
(B) 90, 45, 45
(C) 45, 90, 90
(D) 45, 135, 135
ˆ
ˆ ⋅ dA =
22. If  is a unit vector. The value A
dt
(A) 0
(B) 1
(C)
π
2
(D) π
23. If four non zero vectors satisfy a × b = c × d and
a × c = b × d with a ≠ d and b ≠ c , then
(A) ( a − d ) and ( b − c ) are perpendicular
(B) ( a − d ) and ( b − c ) are parallel
(C) ( a − d ) must equal ( b − c )
(D) ( a − d ) must equal − ( b − c )
24. If two non parallel vectors A and B are equal in mag nitude, then the vectors ( A − B ) and ( A + B ) will be
(A) parallel to each other
(B) parallel but oppositely directed
(C) perpendicular to each other
(D) inclined at an angle θ < 90°
25. A point of application of force F = −5iˆ + 3 ˆj + 2kˆ is
moved from r = 2iˆ + 7 ˆj + 4 kˆ to r = 5iˆ + 2 ˆj + 4 kˆ . The
1
2
work done will be
(A) −22 unit
(B)
(C) −30 unit
(D) +30 unit
22 unit
26. If vectors a = 2iˆ + 4 ˆj − kˆ and b = 3iˆ − 2 ˆj + xkˆ are to be
perpendicular to each other, the value of x should be
−2
(A) 2
(B)
(C) 3
(D) −3
27. When t = 0, a particle at (1, 0, 0), moves towards
(4, 4, 12) with a constant speed of 65 ms −1. The position of the particle is measured in meter and the time
in second. Assume constant velocity, the position of
the particle for t = 2 s is
( 30iˆ − 120 ˆj + 40kˆ ) m
(C) ( 13iˆ − 40 ˆj + 12kˆ ) m
(A)
( 40iˆ + 31ˆj − 120kˆ ) m
(D) ( 31iˆ + 40 ˆj + 120 kˆ ) m
(B)
11/28/2019 6:56:27 PM
3.32 JEE Advanced Physics: Mechanics – I
28. The condition under which vectors ( a + b ) and
( a − b ) should be at right angles to each other is
(A) a ≠ b
(B) a ⋅ b = 0
(C) a = b
(D) a ⋅ b = 1
29. In the arrangement shown in figure, the instantaneous velocities of masses m1 and m2 are v1 and v2 ,
respectively and ∠ACB = 2θ at the instant, then
A
O
θ
(A) y − 2x = 0
(B)
⎛ v ⎞
(C) θ = tan −1 ⎜ 1 ⎟
⎝ 2v2 ⎠
⎛v ⎞
(D) θ = sin −1 ⎜ 1 ⎟
⎝ v2 ⎠
30. One vertex of a parallelopiped is at the point (1, -1, -2)
of rectangular cartesian coordinates. If three adjacent
vertices are at (0, 1, 3), (3, 0, –1) and (1, 4, 1), the volume
of the parallelopiped is
(B) 80 unit
(D) 120 unit
2 2
31. A × B + A ⋅ B is equal to
2
(B)
(A) ( A + B )
(C) y + x = 0
(D) y − x = 0
( A − B )2
5
14
(B)
1
7
(C)
3
14
(D)
1
14
39. A particle is moving in a circle of radius r centred at
O with constant speed v . The change in velocity in
moving from P to Q ( ∠POQ = 40° ) is
v2
Q
40°
v1
P
(B) 30 J
(D) 9 J
(
)
iˆ − 5 ˆj + 2kˆ
and
34. The vector that is added to
ˆ
ˆ
ˆ
3i + 6 j − 7 k to give a unit vector along the x-axis is
03_Vectors_Part 2.indd 32
(B) 15iˆ − 12 ˆj + 13 kˆ
(A)
33. A particle moves from point (1, 0, 2.5) to the point
( − 2, 3, 4) m when a force F = ( iˆ + 4 kˆ ) N acts on it.
)
)
)
38. If a = 2iˆ − 3 ˆj + kˆ and b = 3iˆ + ˆj − 2kˆ , the cosine of angle
θ between them is equal to
(D) A B
32. In equation F = q ( v × B ) , the quantity F
(A) is perpendicular to v only
(B) is perpendicular to B only
(C) is perpendicular to both v and B
(D) is perpendicular to q and B
(
(
(D) 15iˆ + 12 ˆj + 13 kˆ
37. The area of a triangle bounded by vectors a , b and c
is
1 (A)
a+b+c
2
1 (B)
a⋅b + b ⋅c + c ⋅a
2
1 ⎡ (C)
⎣ ( b × c ) + ( c × a ) + ( a × b ) ⎤⎦
6
1 ( a ⋅ b ) + ( b ⋅ c ) + ( c ⋅ a )
(D)
2
2 2
The work done on it is
(A) 6 J
(C) 3 J
2y − x = 0
(C) −15iˆ − 12 ˆj + 13 kˆ
⎛ v ⎞
(B) θ = cos −1 ⎜ 2 ⎟
⎝ 2v1 ⎠
(C) A + B
35. Forces of 1 N and 2 N act along the lines x = 0 and
y = 0 . The equation of the line along which the resultant lies is given by
(A) −15iˆ + 12 ˆj + 13 kˆ
m2
⎛ v ⎞
(A) θ = cos −1 ⎜ 1 ⎟
⎝ 2v2 ⎠
2
(D) 3iˆ + ˆj − 5kˆ
the force in Nm is
m1
2
(C) −3iˆ − ˆj + 5kˆ
(
θ
(A) 400 unit
(C) 40 unit
(B)
36. The radius vector of a point is r = iˆ − 2 ˆj + 3 kˆ m and
a force F = 4iˆ + 5 ˆj acts at that point. The moment of
B
C
iˆ + 3 ˆj + 5kˆ
(A) 3iˆ + ˆj + 5kˆ
(A) 2v cos ( 40° )
(B)
2v sin ( 40° )
(C) 2v cos ( 20° )
(D) 2v sin ( 20° )
11/28/2019 6:56:37 PM
Chapter 3: Vectors
40. Following set of forces act on a body. In which case the
resultant cannot be zero?
(A) 10 N, 10 N, 20 N
(B) 10 N, 10 N, 10 N
(C) 10 N, 20 N, 20 N
(D) 10 N, 20 N, 40 N
41.
d ( )
A×B
dt
dB dA
+B×
(A) A ×
dt
dt
dB dA −
×B
(C) − A ×
dt dt
(B)
dA dB
×B+ A×
dt
dt
(D) 0
42. How many minimum number of coplanar vectors
having different magnitudes can be added to give zero
resultant.
(A) 2
(B) 3
(C) 4
(D) 5
43. How many minimum number of vectors in different
planes can be added to give zero resultant?
(A) 2
(B) 3
(C) 4
(D) 5
44. If none of the vectors A , B and C are zero and if
A × B = 0 and B × C = 0 , the value of A × C is
(A) unity
(B) zero scalar
(C) zero vector
(D) AC cos θ
45. In a watch the average angular velocity is maximum
for the tip of
(A) second’s hand
(B) minute’s hand
(C) hour’s hand
(D) equal for all hands
46. The volume of a parallelopiped bounded by vectors
A , B and C can be obtained from the expression
(A) ( A ⋅ B ) × C
(B) ( A × B ) ⋅ C
(C) ( A ⋅ C ) × B
(D) ( A × B ) × C
47. Pick up the axial vector
(A) force
(C) linear momentum
(B) acceleration
(D) torque
48. Pick out the only vector quantity
(A) pressure
(B) impulse
(C) gravitational potential
(D) coefficient of friction
49. Which of the following quantities is a scalar
(A) electric field
(B) electrostatic potential
(C) magnetic moment
(D) acceleration due to gravity
03_Vectors_Part 2.indd 33
3.33
50. A body of mass 2 kg is constrained to move along
the Y-direction. When a force of 2iˆ + 5 ˆj + 7 kˆ newton
acts on it and the body is displaced through 10 m, the
kinetic energy gained by the body is
(A) 50 J
(B) 100 J
(C) 150 J
(D) 350 J
51. If a denotes a unit vector along an incident light ray,
b a unit vector along refracted ray into a medium
having refractive index x (relative to first medium)
and c a unit vector normal to boundary of two media
and directed towards first medium, then law of refraction is
(B) a × c = x ( c × b )
(A) a ⋅ c = x ( b ⋅ c )
(C) a × c = x ( b × c )
(D) x ( a × c ) = ( b × c )
52. The maximum and minimum magnitudes of the resultant of two given vectors are 17 unit and 7 unit respectively. If these two vectors are at right angles to each
other, the magnitude of their resultant is
(A) 14
(B) 16
(C) 18
(D) 13
53. A vector of magnitude 5 3 unit and another vector
of magnitude 10 unit are inclined to each other at an
angle of 30°. The magnitude of their vector product is
(A) 5 3 unit
(B) 10 unit
(C) 25 3 unit
(D) 50 unit
54. What is the value of linear velocity, if ω = 3iˆ − 4 ˆj + kˆ
ˆ
ˆ
ˆ
and r = 5i − 6 j + 6 k ;
(A) 6iˆ + 2 ˆj − 3 kˆ
(B)
6iˆ − 2 ˆj + 8 kˆ
(C) 4iˆ − 13 ˆj + 6 kˆ
(D) −18iˆ − 13 ˆj + 2kˆ
55. Which (one or more) of the following quantities is a
vector?
(A) Pressure
(B) Power
(C) Current
(D) Angular momentum
56. The
position
vector
of
a
particle
is,
ˆir = ( a c o s ωt ) iˆ + ( a sin ωt ) ˆj . The velocity of the particle is
(A) parallel to position vector
(B) perpendicular to position vector
(C) directed towards the origin
(D) directed away from the origin
57. Given that A = B = C . If A + B = C , then the angle
between A and C is θ1 . If A + B + C = 0 , then the
angle between A and C is θ 2 . What is the relation
between θ1 and θ 2 ?
11/28/2019 6:56:44 PM
3.34 JEE Advanced Physics: Mechanics – I
dr ( 0 ) dx ( 0 ) ˆ dy ( 0 ) ˆ dz ( 0 ) ˆ
v ( 0 ) =
=
i+
j+
k
dt
dt
dt
dt
θ2
2
(A) θ1 = θ 2
(B) θ1 =
(C) θ1 = 2θ 2
(D) None of these
58. The magnitudes of the X and Y components of P
are 7 and 6. Also, the magnitudes of the X and Y
components P + Q are 11 and 9 respectively. What is
the magnitude of Q ?
(A) 5
(C) 8
(B) 6
(D) 9
59. Which of the following statements is false?
(A) Mass, speed and energy are scalar quantities
(B)Momentum, force and torque are vector quantities
(C)Distance is a scalar quantity but displacement is a
vector quantity
(D)A vector has only magnitude, whereas a scalar
has both magnitude and direction
60. The sum of the magnitudes of two vectors is 18 and
the magnitude of their resultant is 12. If the resultant is
perpendicular to one of the vectors, then what are the
magnitudes of the two vectors?
(A) 5, 13
(B) 6, 12
(C) 7, 11
(D) 8, 10
61. When a mass is rotating in a plane about a fixed point
its angular momentum is directed along
(A) the radius
(B) the tangent to the orbit
(C) the axis of rotation
(D) line at an angle of 45° to the axis of rotation
ˆ = 2 ⎡ ( cos t ) iˆ + ( sin t ) ˆj ⎤ .
62. The momentum of a particle isip
⎣
⎦
What is the angle between the force F acting on the
particle and the momentum p ?
(A) 45°
(C) 135°
(B) 90°
(D) 180°
63. Three forces of magnitudes 30, 60 and P newton acting at a point are in equilibrium. If the angle between
the first two is 60°, the value of P is
(A) 25 2
(B)
(C) 30 6
(D) 30 7
64. The
acceleration
of
a
30 3
particle
is
given
by
2π
⎡
⎛ πt ⎞ ⎤
The
initial
cos ⎜ ⎟ kˆ ⎥ ms −2 .
a = ⎢ 2iˆ + 6t ˆj +
⎝ 3⎠ ⎦
9
⎣
conditions
are
r ( 0 ) = x ( 0 ) iˆ + y ( 0 ) ˆj + z ( 0 ) kˆ = 0 ,
v ( 0 ) = 2iˆ + ˆj ms −1 where
2
(
03_Vectors_Part 2.indd 34
)
The position vector at t = 2 s is
( 3iˆ + 8 ˆj + 10kˆ ) m
(C) ( 10iˆ + 3 ˆj + 8 kˆ ) m
(A)
( 8iˆ + 10 ˆj + 3kˆ ) m
(D) ( 3iˆ + 10 ˆj + 8 kˆ ) m
(B)
65. Pick out the only scalar quantity
(A) power
(B) electric field
(C) magnetic moment
(D) average velocity
66. Angular displacement is
(A) a scalar
(B) a vector
(C) either
(D) neither
67. If the vectors P = aiˆ + ajˆ + 3 kˆ and Q = aiˆ − 2 ˆj − kˆ are
perpendicular to each other, then the positive value of
a is
(A) 3
(B) 2
(C) 1
(D) 0
68. A particle of mass m = 5 unit is moving with a uniform speed v = 3 2 unit in the XOY plane along the
line y = x + 4 . The magnitude of the angular momentum of the particle about the origin is
(A) 60 unit
(B)
40 2 unit
(C) zero
(D) 7.5 unit
69. Consider three vectors P , Q and R . Which of the following is independent of choice of coordinate system?
(A) ( P + Q + R )
(B) ( Px + Qx + Rx ) iˆ
(C)
( P iˆ + Q ˆj + R kˆ )
x
y
z
(D) All of these
(
)
70. Force acting on a particle is 2iˆ + 3 ˆj N . Work done by
this force is zero, when a particle is moved on the line
3 y + kx = 5 . Here value of k is
(A) 2
(C) 6
(B) 4
(D) 8
71. A wheel of radius r is rolling on a horizontal surface.
At t = 0 , T is the top most point on the wheel then
magnitude of displacement of point T when wheel
has completed one-fourth rolling is
2r
(B) 2r
(A)
2
⎛π
⎞
(C) r ⎜ − 1 ⎟ + 1
⎝2
⎠
2
⎛π
⎞
(D) r ⎜ + 1 ⎟ + 1
⎝2
⎠
72. The direction of resultant of a and b remains same
when magnitudes of a and b are increased by m
units and n units respectively then
11/28/2019 6:56:52 PM
Chapter 3: Vectors
(A)
a m
=
b n
(B)
a n
=
b m
(C)
a m+n
=
b m−n
(D)
a m−n
=
b m+n
73. Vectors A and B are shown in figure then diagram of
A + B is
B
(A) A + B
(B)
(C)
(D) A + B
74.
(D) 150°
75. If a and b are two unit vectors and R = a + b and also
if R = R , then
(A) R < 0
(B)
R>2
(C) 0 ≤ R ≤ 2
(D) R must be 2
76. If a and b are two sides of a parallelogram and c and
d are the diagonals, then
(A) c 2 + d 2 = a 2 + b 2
B)
c2 + d2 = 2 ( a2 + b 2 )
(D) c 2 − d 2 = 2 ( a 2 − b 2 )
77. In Problem 76, if the angle between a and b is θ ,
then cos θ is equal to
(C) c 2 − d 2 = a 2 − b 2
(A)
(C)
( a2 + b 2 )
4cd
( c2 − d2 )
4 ab
(B)
2iˆ + 2 ˆj – 2kˆ
9
(C)
2 ˆ
i + 2 ˆj − 2kˆ
3
(B)
(
)
(
2 ˆ ˆ ˆ
2i + j + k
3
)
(D) None of these
(D)
a
b
A
(A)
a − 2c
b
(B)
2a − c
b
(C)
2a − c
2b
(D)
a a
−
b 2c
81. For the vectors a and b shown in figure, a = 3iˆ + ˆj
and b = 10 units while θ = 23° , then the value of
R = a + b is nearly
y
b
a
θ
( c2 + d2 )
4 ab
B
c
At what angle should the ball be struck so that it should
rebound from two cushions and go into pocket B?
Assume that in striking the cushion, the ball’s direction of motion changes according to the law of reflection of light from a mirror, i.e., the angle of reflection
equals the angle of incidence.
4cd
)
an instant is ( xiˆ + 2 ˆj − 6 kˆ ) ms −2 , then of x equals
03_Vectors_Part 2.indd 35
(A)
( a2 − b 2 )
78. If magnetic force on a charge is given by q ( V × B )
and a charged particle is projected in a magnetic field
2iˆ + 2 ˆj + 2kˆ tesla. The acceleration of the particle at
(
79. There are two vectors A = 2iˆ + ˆj + kˆ and B = iˆ + 2 ˆj − 2kˆ,
then vector component of A along B is
A+B
The resultant of two forces, one double the other in
magnitude, is perpendicular to the smaller of the two
forces. The angle between the two forces is
(B) 60°
(A) 120°
(C) 90°
(B) 2
(D) 1
80. A billiards ball is at point A on a billiards table whose
dimensions are shown in figure.
A
A+B
(A) 4
(C) 3
3.35
O
(A) 12
(C) 14
x
(B) 13
(D) 15
82. The sum, difference and cross product of two vectors
A and B are mutually perpendicular if
11/28/2019 6:57:01 PM
3.36 JEE Advanced Physics: Mechanics – I
(A)A and B are perpendicular to each other and
A = B
(B) A and B are perpendicular to each other
(C)
A and B are perpendicular but their magnitudes
are arbitrary
(D) A = B and their directions are arbitrary
83. A particle moves in x -y plane. The position vector of
particle at any time t is r = ( 2t ) iˆ + ( 2t 2 ) ˆj m. The
{
}
rate of change of θ at time t = 2 s , where θ is the
angle which its velocity vector makes with positive
x-axis) is
(A)
2
rads −1
17
4
rads −1
7
84. If figure, E equals
(C)
(B)
1
rads −1
14
(D)
6
rads −1
5
B
C
A
E
D
(A) A
(C) A + B
(B)
B
(D) − ( A + B )
85. In figure, D − C equals
E
D
(B) − A
(D) − B
86. In figure, E + D − C equals
B
C
A
D
03_Vectors_Part 2.indd 36
(A)
29 units
(B) 31 units
(C)
37 units
(D)
41 units
88. Which of the following sets of displacements might be
capable of returning a car to its starting point?
(A) 4, 6, 8 and 15 km
(B) 10, 30, 50 and 120 km
(C) 5, 10, 30 and 50 km
(D) 50, 50, 75 and 200 km
89. The magnitude of the vector product of two vectors A
and B may be
(a) Greater than AB
(b) Equal to AB
(c)
(d) Equal to Zero
Less than AB
(A) a, b , c
(B)
(C) a, c , d
(D) a, b , d
b, c, d
90. If A = iˆ + 2 ˆj + 2kˆ and B = 3iˆ + 6 ˆj + 2kˆ , then the vector
in the direction of A and having same magnitude as
B , is
(
)
(B)
7 iˆ + 2 ˆj + 2kˆ
(
)
(D)
7 ˆ
i + 2 ˆj + 2kˆ
9
(A)
7 ˆ
i + 2 ˆj + 2kˆ
3
(C)
3 ˆ
i + 2 ˆj + 2kˆ
7
(
)
(
)
91. The area of the triangle whose vertices are A(1, -1, 2),
B ( 2, 1, − 1 ) and C ( 3 , − 1, 2 ) is
A
(A) A
(C) B
(B)
87. The linear velocity of a rotating body is given by
v = ω × r , where ω is the angular velocity and r
is the radius vector. The angular velocity of a body
ω = ω = iˆ − 2 ˆj + 2kˆ and their radius vector r = 4 ˆj − 3iˆ ,
then v is
B
C
−A
(D) − B
(A) A
(C) B
E
(A) 26
(B)
(C)
(D) 8
13
7 13
92. An engine exerts a force F = 20iˆ − 3 ˆj + 5kˆ N and
moves with velocity v = 6iˆ + 20 ˆj − 3 kˆ ms −1 . The
power of the engine (in watt) is
(A) 45
(B) 75
(C) 20
(D) 10
(
(
)
)
93. The maximum and the minimum magnitudes of the
resultant of two given vectors are 17 unit and 7 unit
respectively. If these two vectors are at right angles to
each other, the magnitude of their resultant is
11/28/2019 6:57:10 PM
3.37
Chapter 3: Vectors
(A) 14
(C) 18
(B) 16
(D) 13
99.
94. For two vectors a and b , if R = a + b and S = a − b , if
2 R = S , when R is perpendicular to a , then
(A)
(C)
a 3
b 7
a
=
b
(B)
1
5
(D)
a
=
b
a
=
b
7
3
5
1
95. Which one is correct?
(A)Resultant of two vectors of unequal magnitude
can be zero
(B)Resultant of three non-coplanar vectors of equal
magnitude can be zero
(C)Resultant of three coplanar vectors is always zero
(D)Minimum number of non-coplanar vectors whose
resultant can be zero is four
96. If r = bt 2iˆ + ct 3 ˆj , where b and c are positive constant, the time at which velocity vector makes an angle
θ = 60° with positive y-axis is
(A)
c
b
(B)
2b
3 3c
(C)
2c
3b
(D)
2b
3c
97. Two vectors A and B are shown in the figure
B
A
(A)
4
1
(B)
2
1
(C)
3
5
(D)
1
4
(
100. If a vector 2iˆ + 3 ˆj + 8 kˆ
(A)
(B)
(C)
(D) None of these
98. Resultant of two vectors A and B is R . Now magnitude of vector is doubled keeping direction same,
then magnitude of the resultant becomes B . Angle
between vector A and B is 120° . Then magnitude of
A is equal to
(B) 2 B
(A) B
B
(C)
(D) 4 B
2
) is perpendicular to the vec-
tor −4iˆ + 4 ˆj + α kˆ , then the value of α is
(A) −1
(C)
1
2
(B)
−1
2
(D) 1
101. The unit vector parallel to the resultant of the vectors
A = 4iˆ + 3 ˆj + 6 kˆ and B = − iˆ + 3 ˆj − 8 kˆ is
(
)
(A)
1 ˆ
3i + 6 ˆj − 2kˆ
7
(C)
1 ˆ
3i + 6 ˆj + 2kˆ
49
(
)
(
)
(B)
1 ˆ
3i + 6 ˆj + 2kˆ
7
(D)
1 ˆ
3i + 6 ˆj − 2kˆ
49
(
)
102. Following sets of three forces act on a body. Whose
resultant cannot be zero?
(A) 10 , 10, 10
(B) 10, 10, 20
(C) 10, 20, 20
(D) 10, 20, 40
103. A vector F1 is along the positive x-axis. If its vec
tor product with another vector F2 is zero then F2
may be
(A) 4 ĵ
Then A − B is given by
03_Vectors_Part 2.indd 37
If R is the resultant of the vectors A and B . The
ratio of minimum value of R to the maximum
1
A
value of R is . Then
may be
4
B
(C)
(
− iˆ + ˆj
(B)
( iˆ + kˆ )
)
(D) −4î
104. The component of 3iˆ + 4 ˆj along iˆ + ˆj
(A)
iˆ + ˆj
2
(C)
5 ˆ ˆ
i+j
2
(
(B)
)
(D)
iˆ
3 ˆ ˆ
(i + j )
2
(
7 ˆ ˆ
i+j
2
)
105. A vector of length l is turned through the angle θ
about its tail. The change in the position vector of its
head is
⎛θ⎞
(A) l sin ⎜ ⎟
⎝ 2⎠
(B)
⎛θ⎞
2l sin ⎜ ⎟
⎝ 2⎠
⎛θ⎞
(C) 2l cos ⎜ ⎟
⎝ 2⎠
⎛θ⎞
(D) l cos ⎜ ⎟
⎝ 2⎠
11/28/2019 6:57:21 PM
3.38 JEE Advanced Physics: Mechanics – I
106. Figure shows three vectors a , b and c , where R is
the mid point of PQ . Then which of the following
relation is correct?
P
a
R
c
Q
b
(A) a + b = 2c
(C) a − b = 2c
a+b =c
(D) a − b = c
(B)
107. A unit vector perpendicular to both vectors iˆ – ˆj + kˆ
and iˆ + ˆj + kˆ is
− iˆ + kˆ
2
(A) iˆ + ˆj
(B)
(C) ˆj + kˆ
iˆ + ˆj
(D)
2
108. The adjacent sides of a parallelogram are represented
by co-initial vectors 2iˆ + 3 ˆj and iˆ + 4 ˆj . The area of
the parallelogram is
(A) 5 units along z-axis
(B) 5 units in x -y plane
(C) 3 units in x -z plane
(D) 3 units in y -z plane
109. Given that A + B + C = 0, out of three vectors two
are equal in magnitude and the magnitude of third
vector is 2 times that of either of the two having
equal magnitude. Then the angles between vectors
are given by
(A) 30°, 60°, 90°
(B)
45°, 45°, 90°
(C) 45°, 60°, 90°
(D) 90°, 135°, 135°
110. If a particle is moving on an elliptical path given by
iˆ r = b cos ( ωt ) iˆ + a sin ( ωt ) ˆj , then radial acceleration
along r is
(A) ω r
(B) ω 2 r
(C) −ω 2 r
(D) None of these
111. Two forces of magnitudes 30, 60 and P Newton acting at a point are in equilibrium. If the angle between
the first two is 60° , the value of P is
(A) 25 2
(B)
(C) 30 6
(D) 30 7
30 3
112. The angle made by the vector 4iˆ − 3 ˆj + 5kˆ with
z-axis is
03_Vectors_Part 2.indd 38
(B) 45°
(A) 30°
(C) 90°
(D) 120°
113. Given A1 = 2 , A2 = 3 and A1 + A2 = 3
then ( A1 + 2 A2 ) ⋅ ( 3 A1 − 4 A2 ) is
(A) −64
(B) 60
(C) −62
(D) None of these
114. Vector A makes equal angles with x , y and
z -axis. Value of its components (in terms of magni
tude of A ) will be
(A)
A
3
(B)
A
2
(C)
3A
(D)
3
A
115. The angle made by the vector A = iˆ + ˆj with x-axis is
(A) 90°
(B)
(C) 22.5°
(D) 30°
45°
116. If a unit vector is represented by 0.5iˆ + 0.8 ˆj + ckˆ , then
the value of c is
(B)
0.11
(A) 1
iˆ
(C)
0.01
(D)
0.39
117. The component of vector A = 2iˆ + 3 ˆj along the vector iˆ + ˆj is
(A)
5
2
(C) 5 2
(B) 10 2
(D) 5
118. There are two force vectors, one of 5 N and other of
12 N at what angle the two vectors be added to get
resultant vector of 17 N , 7 N and 13 N respectively
(A) 0° , 180° and 90°
(B)
0° , 90° and 180°
(C) 0° , 90° and 90°
(D) 180° , 0° and 90°
119. If A = 4iˆ − 3 ˆj and B = 6iˆ + 8 ˆj then magnitude and
direction of A + B will be
⎛ 3⎞
(A) 5, tan −1 ⎜ ⎟
⎝ 4⎠
(B)
⎛ 1⎞
5 5 , tan −1 ⎜ ⎟
⎝ 2⎠
(C) 10, tan −1 ( 5 )
⎛ 3⎞
(D) 25, tan −1 ⎜ ⎟
⎝ 4⎠
120. Two forces 3 N and 2 N are at an angle θ such that
the resultant is R. The first force is now increased to
6 N and the resultant become 2R. The value of θ is
11/28/2019 6:57:35 PM
Chapter 3: Vectors
(A) 30°
(B)
60°
(C) 90°
(D) 120°
121. If A × B = C , then which of the following statements
is wrong?
(A) C ⊥ A
(B) C ⊥ B
(C) C ⊥ ( A + B )
(D) C ⊥ ( A × B )
122. The velocity of a particle is v = 6i + 2j − 2k . The component of the velocity of a particle parallel to vector
a = i + j + k in vector form is
(A) 6iˆ + 2 ˆj + 2kˆ
(B)
(C) iˆ + ˆj + kˆ
(D) 6iˆ + 2 ˆj − 2kˆ
3.39
2iˆ + 2 ˆj + 2kˆ
123. Two forces of magnitude 7 newton and 5 newton act
on a particle at an angle θ to each other. The minimum magnitude of the resultant force is
(A) 5 newton
(B) 8 newton
(C) 12 newton
(D) 2 newton
Multiple Correct Choice Type Questions
This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE is/are correct.
1.
2.
3.
Which of the following will not depend on orientation
of frame of reference
(A) a scalar
(B) a vector
(C) the magnitude of a vector
(D) component of a vector
If P × Q = R , Q × R = P and R × P = Q , then
(A) P , Q and R are coplanar.
(B)angle between P and Q may be less than 90° .
(C) ( P + Q + R ) cannot be equal to zero.
(D) P , Q and R are mutually perpendicular.
5.
Four forces acting on a particle keep it at rest. Then
(A) the forces must be coplanar.
(B) the forces cannot be coplanar.
(C) the forces may or may not be coplanar.
(D)if three of these forces are coplanar, so must be the
fourth.
6.
A vector will not change when
(A) frame of reference is translated
(B) frame of reference is rotated
(C) vector is translated parallel to itself
(D) vector is rotated
If a vector A has magnitude A and n is a unit vector
in the direction of A , then which of the following are
correct
A
A
(B) n =
(A) A = n
A
A
(D) A = An
(C) 1 =
A
If P × Q = R then which of the following statements
is/are correct?
(A) R is perpendicular to ( P + Q )
(B) R is perpendicular to P ⋅ Q
(C) R is perpendicular to P
(D) P is perpendicular to Q × R
7.
A particle is moving towards east with velocity
10 ms −1 . It suddenly turns to left and continues to
move with velocity v now. Change in its velocity Δv
(A) cannot be less than 10 ms −1 in magnitude
4.
(B)will be directed θ NW , where θ will depend
upon magnitude of v
(C)
will be directed θ NE , where θ will depend
upon magnitude of v
(D) None of these
For two vectors A and B which of the following
­relations are not commutative
(B) A − B
(A) A + B
(C) A × B
(D) A ⋅ B
03_Vectors_Part 2.indd 39
8.
11/28/2019 6:57:44 PM
3.40 JEE Advanced Physics: Mechanics – I
9.
Mark the correct statement/s.
(A) the magnitude of the vector 3iˆ + 4 ˆj is 5
(
(B)a force 3iˆ + 4 ˆj
(
)
) N acting on a particle causes a
displacement of 5 ĵ metres. The work done by the
force is 25 J
(C)if a and b represent two adjacent sides of a
parallelogram, a × b gives the area of that
parallelogram
(D)a force has magnitude 20 N. Its component in a
direction making an angle of 60° with the force is
10 3 N
10. The angle that the vector A = 2iˆ + 3 ˆj makes with
y-axis is
⎛ 3⎞
(A) tan −1 ⎜ ⎟
⎝ 2⎠
(B)
⎛ 2⎞
tan −1 ⎜ ⎟
⎝ 3⎠
⎛ 3 ⎞
(C) cos −1 ⎜
⎝ 13 ⎟⎠
⎛ 2⎞
(D) sin −1 ⎜ ⎟
⎝ 3⎠
11. The momentum of a particle is given by
iˆ p = 2 ⎡⎣ ( sin t ) iˆ − ( cos t ) ˆj ⎤⎦ kgms −1 . Select the correct
options
(A)momentum p of the particle is always perpen
dicular to F
(B)momentum p of the particle is always parallel to
F
(C) magnitude of momentum remain constant
(D) None of these
12. Mark the correct statement(s)
(A) displacement is a polar vector.
(B) angular displacement is a polar vector.
(C) displacement is an axial vector.
(D) angular displacement is an axial vector.
Reasoning Based Questions
This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as
Bubble (A) If both statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT 1.
Bubble (B) If both statements are TRUE but STATEMENT-2 is not the correct explanation of STATEMENT 1.
Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.
Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.
5. Statement-1: Vector product of two vectors represents
1. Statement-1: A × B is perpendicular to both A + B
an axial vector.
and A − B .
Statement-2: If v = instantaneous velocity, r = radius
Statement-2: Both A + B and A − B lie in the plane
vector and ω = angular velocity, then ω = v × r .
containing A and B , but A × B lies perpendicular to
6. Statement-1: Minimum number of non-equal vectors
the plane containing A and B.
in a plane required to give zero resultant is three.
ˆ
ˆ
2. Statement-1: Angle between i + j and î is 45° .
Statement-2: If A + B + C = 0 , then they must lie in
one plane.
Statement-2: iˆ + ˆj is equally inclined to both î and ĵ
and the angle between î and ĵ is 90°.
3.
4.
Statement-1: If θ be the angle between A and B ,
A×B
then tan θ = .
A⋅B
Statement-2: A × B is perpendicular to A + B .
Statement-1: If A + B = A − B , then angle between
A and B is 90°.
Statement-2: A + B = B + A .
03_Vectors_Part 2.indd 40
7.
Statement-1: Vector addition of two vectors A and B
is commutative.
Statement-2: A + B = B + A .
8. Statement-1: A ⋅ B = B ⋅ A .
Statement-2: Dot product of two vectors is
commutative.
9. Statement-1: τ = r × F and τ ≠ F × r .
Statement-2: Cross product of vectors is commutative.
11/28/2019 6:57:52 PM
3.41
Chapter 3: Vectors
10. Statement-1: A negative acceleration of a body is associated with a slowing down of a body.
Statement-2: Acceleration is vector quantity.
11. Statement-1: A physical quantity cannot be called as a
vector if its magnitude is zero.
Statement-2: A vector has both, magnitude and
direction.
12. Statement-1: The sum of two vectors can be zero.
Statement-2: The vectors cancel each other, when they
are equal and opposite.
13. Statement-1: Two vectors are said to be like vectors if
they have same direction but different magnitude.
Statement-2: Vector quantities do not have specific
direction.
14. Statement-1: The scalar product of two vectors can be
zero.
Statement-2: If two vectors are perpendicular to each
other, their scalar product will be zero.
15. Statement-1: Multiplying any vector by a scalar is a
meaningful operations.
Statement-2: Taking dot product of a scalar and a vector is meaningless.
16. Statement-1: A null vector is a vector whose magnitude is zero and direction is arbitrary.
Statement-2: A null vector does not exist.
17. Statement-1: If dot product and cross product of A
and B are zero, it implies that either of the vectors A
and B must be a null vector.
Statement-2: Null vector is a vector with zero
magnitude.
18. Statement-1: The cross product of a vector with itself
is a null vector.
Statement-2: The cross product of two vectors results
in a vector quantity.
19. Statement-1: The minimum number of non coplanar
vectors whose sum can be zero, is four.
Statement-2: The resultant of two vectors of unequal
magnitude can be zero.
20. Statement-1: If A ⋅ B = B ⋅ C , then A may not always
be equal to C .
Statement-2: The dot product of two vectors involves
cosine of the angle between the two vectors.
Linked Comprehension Type Questions
This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a
Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct.
(For the sake of competitiveness there may be a few questions that may have more than one correct options)
Comprehension 1
4.
Consider a parallelogram whose adjacent sides are given
by a = 2m + n and b = m − 2n , where m and n are unit
vectors inclined at an angle of 60° . Based on this information, answer the following questions.
1.
2.
3.
5.
The length of the diagonal found in PROBLEM 1 is
(A)
7 units
(B)
10 units
(C)
12 units
(D)
13 units
03_Vectors_Part 2.indd 41
(A)
7 units
(B)
10 units
(C)
12 units
(D)
13 units
The area of the parallelogram is
(A) 2 3 sq. units
One diagonal of the parallelogram is represented by
the vector
(B) 3 m + n
(A) m + n
(C) 3 m − n
(D) − m + 2n
The other diagonal of the parallelogram is represented
by the vector
(B) m + 3n
(A) n − 3 m
(C) − m − 3n
(D) m − n
The length of the diagonal found in PROBLEM 2 is
6.
(B)
5 3 sq. units
(C)
5
3 sq. units
2
(D)
5
3 sq. units
4
If the angle between the diagonals is θ . Then
⎛ 2 ⎞
(A) θ = cos −1 ⎜
⎝ 91 ⎟⎠
⎛ 1 ⎞
(B) θ = cos −1 ⎜
⎝ 91 ⎟⎠
⎛ 7 ⎞
(C) θ = cos −1 ⎜
⎝ 91 ⎟⎠
⎛ 4 ⎞
(D) θ = cos −1 ⎜
⎝ 91 ⎟⎠
11/28/2019 6:57:57 PM
3.42 JEE Advanced Physics: Mechanics – I
Comprehension 2
13.
One of the principles of addition of vector is the parallel
ogram law which says that “if two vectors a and b are
acting at point O , then the resultant of these two vectors
is obtained by completing
a parallelogram whose adjacent
sides are a and b . The resultant is the diagonal of the parallelogram passing through the same point O .” Based on
this information, answer the following questions.
7. The diagonal p of parallelogram passing through
point O is
(A) a − b
(B) a + b
(C) a + 2b
(D) b + 2 a
8.
F1 is written as
( 5iˆ − 5 ˆj ) N
(C) ( 5iˆ + 5 ˆj ) N
Comprehension 4
Four vectors are shown in the figure where A = 5 2 m,
B = 10 m, C = 10 m and D = 10 m. Based on this information, answer the following questions.
Y
The diagonals p and q will be perpendicular to each
other if
(A) a = b
(B) a = 2 b
(C) b = 2 a
(D) None of these
10. Angle between a + b and a × b is
(A) 0°
(B)
(C) 60°
(D) 90°
45°
Comprehension 3
Three forces F1 , F2 and F3 act on a particle as shown in the
figure. If F1 = 5 2 N , F2 = 10 N and F3 = 20 N. Based
on this information, answer the following questions.
Y
C
14.
16.
(A) zero
(B) 10 m
(C) −5 m
(D) −5 3 m
( Ay + By + Cy + Dy ) is equal to
(A) 5 ( 1 − 3 ) m
(B)
(C) 5 3 m
(D) 10 ( 3 − 1 ) m
Three forces F1 , F2 and F3 are acting on a body as shown
in figure.
y
30
°
12.
5 ( −1 ) m
Comprehension 5
F2
120°
F1
F3
11.
(B) −10 3 m
(D) 10 m
( Ax + Cx ) is equal to
X
60°
D
(A) 20 3 m
(C) zero
15.
X
( Bx + Dx ) is equal to
F3
45°
45°
30°
30°
60°
F1
F2
A
B
The other diagonal q of the parallelogram is
(A) a − b
(B) a + b
(C) a + 2b
(D) b + 2 a
9.
( −5iˆ + 5 ˆj ) N
(D) 5 2 ( iˆ + ˆj ) N
(B)
(A)
X component of F2 is
x
If F1 = 10 N , F2 = 10 3 N and F3 = 20 N .
(A) 5 N
(B)
−5 N
Based on this information, answer the following questions.
(C) 5 3 N
(D) −5 3 N
17. External force required to keep block at rest is (in
newton)
(A) 10 3 N
(B)
(C) −10 N
(D) 10 N
Y component of F3 is
03_Vectors_Part 2.indd 42
−10 3 N
−20 3 ĵ
(A) 20 3 ĵ
(B)
(C) 20 2î
(D) 20 iˆ − 20 3 ˆj
11/28/2019 6:58:09 PM
3.43
Chapter 3: Vectors
18.
a + b − c = _____
(A) −2 a
(C) −2c
21. ( a × b ) ⋅ c = _____
F1 + F2 − F3 is given by
20.
(A) 20 newton in x-direction
(B) 20 newton in y-direction
(C) 20 3 newton in y-direction
(D) −20 newton in x-direction
19. F2 + F3 − F1 is given by
(A) abc
(C) −1
22. a ⋅ b + b ⋅ c = _____
(A) 20 newton at angle 120° with x-axis
(B)
40 newton at angle 120° with x-axis
(B)
−2b
(D) zero
(B)
−abc
(D) 0
(C) 20 3 newton at angle 150° with x-axis
(A) −b 2
(B)
−a 2
(D) 40 3 newton at angle 150° with x-axis
(C) −c 2
(D) zero
23. If triangle is equilateral, then angle between a and b
is
Comprehension 6
Three vectors a , b , c form a triangle taken in same order.
Based on this information, answer the following questions.
(A) 60°
(B) 120°
(C) 150°
(D) 90°
b
a
c
Matrix Match/Column Match Type Questions
Each question in this section contains statements given in two columns, which have to be matched. The statements in
COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct ­darkening of
bubbles will look like the following:
A
B
C
D
1.
q
q
q
q
q
r
s
t
r
r
r
r
s
s
s
s
t
t
t
t
2.
Match the following
Column-I
(A) A ⋅ B
(B) ( A ⋅ B ) C
(p) Defined
(C) ( A ⋅ B ) ⋅ C
(r) Scalar
(D) ( A × B ) × C
(s) Vector
03_Vectors_Part 2.indd 43
p
p
p
p
p
Column-II
(q) Not Defined
Match the following
Column-I
(A) A × B
(B) ( A × B ) ⋅ A
(C) A × A
(D) 0 × A
Column-II
(p) Zero Scalar
(q) Perpendicular to A
(r) Zero Vector
(s) Perpendicular to B
11/28/2019 6:58:15 PM
3.44 JEE Advanced Physics: Mechanics – I
3.
6.
Match the following
Column-I
Column-II
(A) Parallel Vectors
(p)
(B) Perpendicular Vectors
⎛ A⋅B⎞ (q) ⎜ 2 ⎟ B
⎝ B ⎠
(C) A × B + C × A = 0
(D) Area of Parallelogram
(E) Component Vector of A
along B
4.
Column-I
Column-II
(A) Resultant of two
vectors of equal
magnitude
dA
=0
(p) A ⋅
dt
(r) A × B = 0
(B) For a vector A of
fixed magnitude
and direction
(q) will be zero
(s) B = C + kA
(C) For any vector A
A×B
(r) lies along the angle
bisector of two
vectors
(t) A ⋅ B = 0
(D) Resultant of n
vectors of equal
magnitude each
inclined at an
2π
angle
with the
n
preceding vector
Match the following
Column-I
Column-II
(A) Resultant of two ordered
vectors
(p) Zero Vector
(B) Dot Product
(q) Always
Negative
7.
dA (s) A ×
=0
dt
If R = a + b and S = a − b also θ is angle between a
and b .
(C) A ⋅ B = 0 and A × B = 0
(r) Triangle Law
Column-I
Column-II
(D) A × ( B × C ) + B × ( A × C )
+C × ( A × B )
(s) Commutative
in nature
(A) R2 + S2
(p) R is perpendicular to a
(B) R2 − S2
(q) 2 ( a 2 + b 2 )
(E) Three
vectors A, B and
C taken in same order
form a closed polygon,
then A ⋅ B + B ⋅ C + C ⋅ A is
5.
Match the following
(t) A = 0 OR
B =0
(C)
(r) 4 a ⋅ b
R
S
⎛θ⎞
(s) tan ⎜ ⎟ if a = b
⎝ 2⎠
(D) R < S
Match the following
Column-I
(A)
A+B = A−B
Column-II
(p)
A = B
(B) A − B = A + B
(q) 2 ( B × A )
(C) ( A + B ) × ( A − B )
(r) B = 0
(D) ( A + B ) ⊥ ( A − B )
(s) A ⊥ B
(t) A ⋅ B = 0
03_Vectors_Part 2.indd 44
8.
Three coplanar forces F1 , F2 and F3 are acting simultaneously on a particle. COLUMN-I contains different
operation between forces and COLUMN-II contains
their magnitude. Match them
F1
F1 = 10 N
F2 = 10 N
F2
F3 = 10 √2 N
135°
F3
11/28/2019 6:58:23 PM
Chapter 3: Vectors
9.
Column-I
Column-II
Column-I
Column-II
(A) F1 + F2 + F3
(p) 0
(B)
(q) b − c = a
(B) F1 + F2 − F3
(q) 20
(C) F1 − F2 + F3
(r) 20 2
(D) F3 ⋅ ( F1 + F2 )
(s) 200
(C)
A×B
(p)
(B)
A−B
(q) 6
(C) A ⋅ B
(D)
A+B
(D)
(A)
c
(s) a + b = c
b
a
11
11. In vector algebra match the vector operations with
conditions under which operation is possible.
(r)
35
Column-I
Column-II
(s)
90
(A) A + B = A − B
(p) Not possible
(B)
A+B = A−B
(q) θ = 90°
(C)
A × B = A⋅C
(r) B = 0
(D) A × B = A ⋅ C
Column-II
b
(r) a + b = − c
b
c
10. COLUMN-I contains vector diagram of three vectors
a , b , c and COLUMN-II contains vector equation.
Match them
Column-I
a
c
Column-II
(A)
b
a
If A = 2iˆ + 3 ˆj − kˆ and B = iˆ + 2 ˆj + 2kˆ then
Column-I
c
3.45
(s) θ = 45°
(p) a − ( b + c ) = 0
a
(Continued)
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data
given in the question(s).
1. A particle has a displacement of 12 m towards east,
4. Two vectors A and B are such that A + B = C .
3 m towards north and 4 m vertically upwards. Find
Also A 2 + B2 = C 2. Find the angle between A and B,
the resultant magnitude of these displacements. Give
in degree.
your answer in cm.
5. Given that v1 + v2 = v1 − v2 . Calculate the angle
2. At what angle, in degree, two forces of magnitude
between v1 and v2 , in degree.
A + B and A − B ( A > B ) act, so that their resultant is
6. The resultant of two forces F1 and F2 is P. If F2 is
2
2
3A + B ?
reversed, then the new resultant is Q . Find the value
2
2
3. Two equal vectors act at a point. The square of their
of P + Q .
2
2
F1 + F2
resultant is 3 times their product. Find the angle, in
degree, between the vectors.
03_Vectors_Part 2.indd 45
11/28/2019 6:58:30 PM
3.46 JEE Advanced Physics: Mechanics – I
7.
8.
9.
Calculate the x and y components of a vector in
first quadrant having magnitude same as that of
A = 7 iˆ + 24 ˆj and parallel to B = 3iˆ + 4 ˆj .
10. One vertex of a parallelopiped is at the point
( 1, − 1, − 2 ) of rectangular Cartesian coordinates. If
three adjacent vertices are at ( 0 , 1, 3 ) , ( 3 , 0 , − 1 ) and
( 1, 4, 1 ). Calculate the volume of the parallelopiped.
Find ( y − x ) such that the vectors a = xiˆ + 4 ˆj − 8 kˆ and
b = 2iˆ + yjˆ + 4 kˆ are collinear.
If a vector P makes angles α , β and γ with respect to
X , Y and Z axes of a rectangular coordinate system,
then find the value of sin 2 α + sin 2 β + sin 2 γ .
ARCHIVE: JEE MAIN
1.
[Online April 2019]
Let A1 = 3 , A2 = 5 and A1 + A2 = 5 . The value of
( 2A1 + 3 A2 ) ⋅ ( 3 A1 − 2A2 ) is
(A) −106.5
(B)
−118.5
(C) −99.5
(D) -112.5
2. [Online January 2019]
In the cube of side a shown in the figure, the vector
from the central point of the face ABOD to the central
point of the face BEFO will be
B
A
5.
D
y
(C)
(
)
(B)
(
)
(D)
1 ( ˆ ˆ)
a k−i
2
1 ˆ ˆ
a j−k
2
3.
[Online January 2019]
Two vectors A and B have equal magnitudes. The
magnitude of ( A + B ) is n times the magnitude of
( A − B ) . The angle between A and B is
⎛ n2 − 1 ⎞
cos −1 ⎜ 2
⎝ n + 1 ⎟⎠
⎛ n − 1⎞
(A) cos −1 ⎜
⎝ n + 1 ⎟⎠
(B)
⎛ n − 1⎞
(C) sin −1 ⎜
⎝ n + 1 ⎟⎠
⎛ n2 − 1 ⎞
(D) sin −1 ⎜ 2
⎝ n + 1 ⎠⎟
03_Vectors_Part 2.indd 46
[Online January 2019]
(B) Zero
(C) 10 3 ms
6.
a
1 (ˆ ˆ )
a i −k
2
1 ˆ ˆ
(A)
a j −i
2
(D) 120°
(A) 10 ms −1
a
x
(C) 60°
90°
the change in velocity of the particle, when it moves
through an angle of 60° around the centre of the
circle?
a
F
(B)
stant speed of 10 ms −1 . What is the magnitude of
E
O
(A) 30°
A particle is moving along a circular path with a con-
z
H
G
4. [Online January 2019]
Two forces P and Q , of magnitude 2F and 3F ,
respectively, are at an angle θ with each other. If the
force Q is doubled, then their resultant also gets doubled. Then, the angle θ is
−1
(D) 10 2 ms −1
[Online 2018]
Let A = iˆ + ˆj and B = 2iˆ − ˆj . The magnitude of a
coplanar vector such that A ⋅ C = B ⋅ C = A ⋅ B , is given
by
(
)
(
)
(A)
20
9
(B)
5
9
(C)
9
12
(D)
10
9
7. [Online 2015]
A vector A is rotated by a small angle Δθ radians
( Δθ 1 ) to get a new vector B. In that case B − A is
(A) 0
(C)
A Δθ
(B)
(D)
⎛
Δθ 2 ⎞
A ⎜1−
⎟
⎝
2 ⎠
B Δθ − A
11/28/2019 6:58:41 PM
Chapter 3: Vectors
3.47
ARCHIVE: JEE Advanced
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical
value obtained after series of calculations based on the data
provided in the question(s).
π
rads −1 . If A + B = 3 A − B at time t = τ for
6
the first time, the value of τ , in seconds, is ______.
ω=
1.
[JEE (Advanced) 2018]
Two vectors A and B are defined as A = aiˆ and
B = a cos ωtiˆ + sin ωtjˆ , where a is a constant and
iˆ
03_Vectors_Part 2.indd 47
(
)
11/28/2019 6:58:43 PM
3.48 JEE Advanced Physics: Mechanics – I
Answer Keys—Test Your Concepts and Practice Exercises
Test Your Concepts-I (Based on Addition, Subtraction and Resolution)
1. No, Yes
2. (a) Bx = 6 and By = 3
(b) 3 5 m
1
(c) tan −1 ⎛⎜ ⎞⎟ , with x-axis
⎝ 2⎠
3.
4.
A
B
A+B
A−B
B−A
Magnitude
5
10
5 5
5 5
5 5
Direction
(q with x-axis)
⎛ 3⎞
tan −1 ⎜ ⎟
⎝ 4⎠
⎛ 4⎞
tan −1 ⎜ ⎟
⎝ 3⎠
⎛ 1⎞
tan −1 ⎜ ⎟
⎝ 2⎠
⎛ 11 ⎞
tan −1 ⎜ ⎟
⎝ 2⎠
⎛ 11 ⎞
tan −1 ⎜ ⎟
⎝ 2⎠
Below x-axis
Above x-axis
Above x-axis
Below negative
x-axis
Above x-axis
Test Your Concepts-III (Based on Cross
Product, Scalar and Vector Triple Product)
205 m
7. 8iˆ − 2 ˆj ; 2 17 kgf
8. a and b are parallel
1. (a) −21iˆ + 29 ˆj + 41kˆ
10. 2v sin ( 20° )
(b) −23iˆ + 29 j + 38 kˆ
13. Yes
2. iˆ ( Ay Bz − Az By ) − ˆj ( Ax Bz − Az Bx ) + kˆ ( Ax By − Ay Bx )
Test Your Concepts-II (Based on Dot Product)
1. Ax Bx + Ay By + Az Bz
2. 15°
⎛ 4
6
3 ⎞ ⎛
2
5
3. (a) A ⎜
,
,−
, −
,
⎟ , B ⎝⎜ −
⎝ 61 61
⎠
61
78
78
(b) 31.3°
5.
1
234 sq units
2
5. Area, with direction as the outward normal.
6. Yes
7.
7 mv02
, along negative z-axis.
20 g
8. v =
44
142
−1 ⎛ 12 ⎞
6. 12; cos ⎜
⎝ 77 ⎟⎠
8.
7 ⎞
⎟
78 ⎠
4.
(
E
, along +x axis or −x axis
B
9. 4 ˆj − kˆ
)
177
J
7
03_Vectors_Part 2.indd 48
11/28/2019 6:58:49 PM
3.49
Chapter 3: Vectors
Single Correct Choice Type Questions
1. A
2. A
3. D
4. B
5. D
6. A
7. C
8. C
9. A
10. C
11. A
12. C
13. A
14. A
15. B
16. D
17. C
18. D
19. A
20. C
21. A
22. A
23. B
24. C
25. C
26. B
27. D
28. C
29. B
30. C
31. D
32. C
33. C
34. C
35. B
36. A
37. C
38. D
39. D
40. D
41. B
42. B
43. C
44. C
45. A
46. B
47. D
48. B
49. B
50. A
51. C
52. D
53. C
54. D
55. D
56. B
57. B
58. A
59. D
60. A
61. C
62. B
63. D
64. B
65. A
66. B
67. A
68. A
69. A
70. A
71. D
72. A
73. B
74. A
75. C
76. B
77. C
78. A
79. A
80. C
81. A
82. D
83. A
84. D
85. A
86. D
87. A
88. A
89. B
90. A
91. C
92. A
93. D
94. A
95. D
96. C
97. D
98. C
99. C
100. C
101. A
102. D
103. D
104. D
105. B
106. A
107. B
108. A
109. D
110. C
111. D
112. B
113. A
114. A
115. B
116. B
117. A
118. A
119. B
120. D
121. D
122. B
123. D
Multiple Correct Choice Type Questions
1. A, B, C
2. C, D
3. B, C
4. B, C
5. C, D
6. A, B, C
7. B, C, D
8. A, C
9. A, C
10. B, C
11. A, C
12. A, D
Reasoning Based Questions
1. A
2. A
3. B
4. B
5. C
6. B
7. B
8. A
9. C
10. B
11. D
12. A
13. C
14. A
15. B
16. C
17. B
18. B
19. C
20. A
Linked Comprehension Type Questions
1. C
2. B
3. A
4. D
5. C
6. D
7. B
8. A
9. A
10. D
11. B
12. B
13. C
14. C
15. A
16. A
17. B
18. A
19. B
20. C
21. D
22. A
23. B
Matrix Match/Column Match Type Questions
1. A → (p, r)
B → (p, s)
C → (q)
D → (p, q, s)
2. A → (q, s)
B → (p)
C → (r)
D → (r)
3. A → (r)
B → (t)
C → (s)
D → (p)
E → (q)
4. A → (r, s)
B → (s)
C → (t)
D → (p)
E → (q)
5. A → (s, t)
B → (r)
C → (q)
D → (p)
6. A → (r)
B → (p)
C → (s)
D → (p)
7. A → (q)
B → (r)
C → (s)
D → (p)
8. A → (p)
B → (r)
C → (q)
D → (s)
9. A → (s)
B → (p)
C → (q)
D → (r)
10. A → (r)
B → (s)
C → (p)
D → (q)
11. A → (r)
B → (q)
C → (s)
D → (p)
03_Vectors_Part 2.indd 49
11/28/2019 6:58:49 PM
3.50 JEE Advanced Physics: Mechanics – I
Integer/Numerical Answer Type Questions
1. 1300
2. 60
3. 60
4. 90
5. 90
6. 2
7. x = 15 and y = 20
8. 2
9. 2
10. 40
3. B
5. A
ARCHIVE: JEE MAIN
1. B
2. A
4. D
6. B
7. C
ARCHIVE: JEE advanced
Integer/Numerical Answer Type Questions
1. 2
03_Vectors_Part 2.indd 50
11/28/2019 6:58:49 PM
CHAPTER
4
Kinematics I
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Rest, Motion and Position
(i) Vertical Motion Under Gravity
(b) Distance Displacement
( j) Motion in a Plane
(c) Average Speed and Average Velocity
(k) Relative Motion in One Dimension
(d) Instantaneous Speed and Instantaneous
(l) Relative Motion in Two Dimensions
Velocity
(m) Distance of Closest Approach between
(e) Average and Instantaneous Acceleration
Moving Bodies
(f) Uniformly Acceleration Motion
(n) River-Swimmer Problems
(g) Variable Accelerated Motion
(o) Aeroplane-Wind Problems
(h) Graphical Interpretation and Graphs
(p) Rain-Man-Wind Problems
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
and Advanced) are also given.
RECTILINEAR MOTION AND MOTION uNDER GRAvITY
INTRODuCTION TO CLAssICAL
MEChANICs
The branch of Physics dealing with motion of particles
or bodies in space and time is called Mechanics. As
long as the velocity of the moving bodies is small
in comparison to the velocity of light (c), the linear
dimensions and the time intervals remain invariable
in all Reference Frames (platform(s) from where
motion is being observed), i.e., they do not depend
on choice of reference frame. Mechanics dealing with
such like motion (also called as Non-Relativistic
motion) is called as Classical Mechanics. However,
when the bodies move with speeds comparable to
the speed of light (called as relativistic speeds), then
the part of Physics dealing with such like motion(s)
is called Relativistic Mechanics. An interesting fact
04_Kinematics 1_Part 1.indd 1
about relativistic mechanics is that it is more general
and reduces to classical mechanics for the case of small
(non relativistic) velocities. In this chapter, we shall be
describing and studying motion in terms of space and
time while ignoring the causes that produce motion.
This particular part of Classical Mechanics is called
Kinematics. Furthermore, in this part of chapter,
we shall be limiting ourselves to the motion in one
dimension and two dimensions i.e., motion along a
straight line, also called as Rectilinear Motion and
planar motion. From our everyday experience, we
observe that actually motion represents a continuous
change in the position of an object. In Physics, we can
divide motion into three categories
(a) Translational Motion (studying now).
EXAMPLE: A car moving down a highway.
11/28/2019 7:01:48 PM
4.2 JEE Advanced Physics: Mechanics – I
(b) Rotational Motion (to study in Rotational
Dynamics).
EXAMPLE: Spinning i.e., rotation of earth about its
own axis.
(c) Vibrational Motion (to study in Simple Harmonic
Motion).
EXAMPLE: Back and forth (or to and fro) motion of
a pendulum (or a spring).
In the Chapters to came, we shall be discussing the
branch of Classical Mechanics called DYNAMICS,
where we shall be studying the motion along with
the cause p
­ roducing it.
CONCEPT OF POINT OBJECT
(PARTICLE MODEL)
In our study of translational motion, we shall be
using the concept of Point Object also called as the
Particle Model. A body is considered as a point object
depending upon the nature of the motion followed
by the body. In general, an object is regarded a pointobject when it travels large distances in comparison
to its own size and dimensions.
Also, in planetary motion, the bodies under
consideration can be regarded as point objects when
distances of separation are very large.
Example:
The planets revolving around the sun may be considered
as point objects.
CONCEPT OF REFERENCE FRAME
A reference frame is a platform from where a physical phenomena, such as motion, is being observed.
Reference frames are mainly of two types
(a) Inertial (Non-Accelerated frames)
(b) Non-Inertial (Accelerated frames)
The detailed discussion about both these frames follows in the next chapter. As of now, for this ­chapter,
the reference frame is a frame (fixed or moving with
constant ­velocity) i.e., non accelerating frame. The
best convenient reference frame for this chapter
would be the Ground Frame. Please note here, that
I said “the best convenient” which has convenience,
but not accuracy attached to it. For practical purposes, we regard earth as an inertial frame (though
04_Kinematics 1_Part 1.indd 2
it is accelerating), hence I said convenient and not
accurate.
State of Rest and Motion
A particle is said to be in the state of rest when it does
not change its position w.r.t. s­ urroundings with the
passage of time.
A particle is said to be in the state of motion
when it changes its position w.r.t. surroundings with
the passage of time.
Rest and Motion, these two terms are not absolute i.e., complete in themselves. These terms are relative terms i.e., a body can be in the state of rest and
motion simultaneously, depending upon the relative
observer viewing motion.
Example:
Four persons sitting in the moving car are at rest w.r.t. an
observer sitting in the car, whereas the same four persons
are in motion w.r.t. a stationary observer viewing them
from the ground.
Position of a Particle
The position of a particle is the location of the particle in space at a certain moment of time. In OneDimensional Motion the position of the particle is
denoted by x or y or z . However in 2-Dimensional
and 3-Dimensional motion it is denoted by r , where
r = xiˆ + yjˆ + zkˆ
Distance and Displacement
(Relative Position Vector)
The length of the actual path followed between the
initial and the final points in the motion is called
Distance. Its SI unit is metre and cgs unit is cm.
y
Distance path
i
ri
Dis
pla
ce
Δr = rf – ri
me
nt
f
rf
x
11/28/2019 7:01:50 PM
4.3
Chapter 4: Kinematics I
The change in position vector of a particle going
from initial position (say i) to final position ( say f ) is
called Displacement or Relative Position Vector of
the particle.
It is denoted by Δr , such that
Δr = r f − ri
Illustration 1
Ram takes path 1 (straight line) to go from P to Q
and Shyam takes path 2 (semicircle).
(a) Find the distance travelled by Ram and Shyam.
(b) Find the displacement of Ram and Shyam.
2
Conceptual Note(s)
P
In 2-Dimensional or 3-Dimensional motion, the
­displacement is denoted by Δr = rf − ri
Properties of Displacement
(a) It is a vector quantity.
(b) It has units same as that of distance.
(c) It is independent of the choice of origin.
(d) It is unique (one and only one) for any kind of
motion between two points.
(e) It is always concealing about the actual track followed by the particle’s motion between any two
points.
(f) It can be positive, negative and even be zero.
(g) The magnitude of the displacement is always less
than or equal to the distance for particle’s motion
between two points i.e.,
0 ≤ Displacement ≤ Distance
(h) A body may have finite distance travelled for
zero displacement.
Conceptual Note(s)
(a) For Rectilinear (straight line) motion, always make
sure that the selected Cartesian Co-ordinate axis
coincides with the line followed by the particle.
(b) In other words Displacement may also be defined
as the vector drawn from the initial point to the
final point in the motion of the particle.
(c) Please keep in mind that distance may or may not
be equal to the magnitude of the displacement.
Displacement
≤1
(d) In general, we have 0 ≤
Distance
04_Kinematics 1_Part 1.indd 3
Q
1
100 m
In rectilinear motion of a particle, the displacement is
denoted by Δx = s = x final − xinitial
Solution
(a) Distance travelled by Ram = 100 m
Distance travelled by Shyam = π ( 50 m ) = 50π m
(b) Displacement of Ram = 100 m
Displacement of Shyam = 100 m
Illustration 2
The position x (in metre) of a particle varies with
time t (in second) as t = x + 3 . Calculate the
(a) displacement of the particle from t = 0 to t = 3 s
(b) displacement of the particle from t = 3 to t = 6 s
(c) distance travelled and displacement from t = 0
to t = 6 s
Solution
Since t = x + 3
2
x = ( t − 3 ) …(1)
⇒
So, we get x as a function of t . Let us draw a table for
x and t from t = 0 to t = 6 . Then we get
t
0
1
2
3
4
5
6
x
9
4
1
0
1
4
9
This table gives us the answers to the three parts.
So, for (a), we have Δx1 = −9 m
For (b), we have Δx2 = +9 m
For (c), we have,
Distance = −9 m + 9 m = 18 m and
Displacement = zero.
Δ x1 = –9 m
Δ x2 = +9 m
0
1
4
9
x
11/28/2019 7:01:55 PM
4.4 JEE Advanced Physics: Mechanics – I
AVERAGE SPEED AND AVERAGE VELOCITY
vav =
⇒
⎛t⎞
⎛t⎞
⎛t⎞
⎛t⎞
v1 ⎜ ⎟ + v2 ⎜ ⎟ + v3 ⎜ ⎟ + ... + vn ⎜ ⎟
⎝ n⎠
⎝ n⎠
⎝ n⎠
⎝ n⎠
vav =
t
⇒
vav =
Average speed of a particle, in a given interval of time
is the ratio of the total distance travelled by the particle to the total time taken. It is a scalar quantity,
vav =
Total distance travelled s1 + s2 + s3 + ...
=
Total time taken
t1 + t2 + t3 + ...
Average velocity of the particle is the ratio of the
­displacement of the particle to the time taken. It is a
vector quantity.
In one dimensional motion, we have
Total Displacement Δx x f − xi
vav =
=
=
t f − ti
Total Time
Δt
Please note that, average speed may or may not be
equal to the magnitude of average velocity. So,
Average Speed ≠ vav (sometimes may be equal)
So, we observe that when an interval is divided
into n equal time parts, then the “average speed is
the ­simple average of the speeds in the respective
intervals”.
Consider an interval of length l , divided equally in
n equal length parts. Let a particle travel first length
part with a velocity v1 , second with a velocity v2
and so on the nth length part with a velocity vn
(as shown). Let t1 , t2 , …, tn be the respective time
taken by the particle to cover the respective interval.
Then
Concept of Average Speed
Average Speed =
Average speed is simply defined as the total distance
travelled by a body per unit total time taken to complete the motion.
A
Total Distance Travelled
Average Speed =
Total Time Taken
For Interval Divided in Equal Time Parts
Let a particle go from A to B in a time t (say). Now
t
if we say that for time , the particle had a velocn
t
ity v1 , for next
it had a velocity v2 and for the
n
nth such equal time interval it had a velocity vn (as
shown). Let the corresponding distances travelled in
each respective interval be l1 , l2 , l3 , …., ln . Then
A
04_Kinematics 1_Part 1.indd 4
v1 + v2 + v3 + ... + vn
n
For Interval Divided in Equal Length Parts
In two or three dimensional motion, we have
Total Displacement Δr r f − ri
vav =
=
=
Total Time
Δt t f − ti
Average Speed =
l1 + l2 + l3 + ... + ln
t t t
t
+ + + ... + ( n times )
n n n
n
⇒
Total Distance Travelled
Total Time Taken
l1
l2
l3
ln
v1
t
n
v2
t
n
v3
t
n
vn
t
n
B
⇒
⇒
⇒
Total Distance Travelled
Total Time Taken
l
n
l
n
l
n
l
n
l
n
t1
v1
t2
v2
t3
v3
t4
v4
tn
vn
l l
+ + ...n times
l
vav = n n
=
t1 + t2 + ... + tn
t1 + t2 + ... + tn
vav =
B
l
l
l
l
+
+ ... +
nv1 nv2
nvn 1
1
1
+ + ... +
v1 v2
vn
1
=
vav
n
So, we observe that when an interval is divided into
n equal length parts, then the “reciprocal of the average speed is equal to the average of the reciprocals
of the speeds in respective intervals”.
11/28/2019 7:02:00 PM
Chapter 4: Kinematics I
Problem Solving Technique(s)
vav =
So, we observe that
(a) when an interval is divided into n equal time parts,
then the “average speed is the simple average of
the speeds in the respective intervals”.
(b) when an interval is divided into n equal length
parts, then the “reciprocal of the average speed
is equal to the average of the reciprocals of the
speeds in respective intervals”.
(c) Time Average Speed: When particle moves with
different uniform speed v1, v2, v3 … etc. in different
time intervals t1, t2, t3, … etc. respectively, its average speed over the total time of journey is given
as
Total distance covered
vav =
Total time elapsed
vav =
d1 + d2 + d3 + ...... v1t1 + v2 t2 + v3t3 + ......
=
t1 + t2 + t3 + ......
t1 + t2 + t3 + ......
(d) Distance Average Speed: When a particle
describes different distances d1, d2, d3, … with different time intervals t1, t2, t3, … with speeds v1, v2,
v3, … respectively, then the speed of particle averaged over the total distance can be given as
Total distance covered d1 + d2 + d3 + .......
vav =
=
Total time elapsed
t1 + t2 + t3 + ......
d + d + d3 + ......
vav = 1 2
d1 d2 d3
+
+ + ......
v1 v2 v3
(e) If speed is continuously changing with time then
vav =
∫ vdt
∫ dt
(f) When a particle moves with a speed v1 for half the
time and with a speed v2 for the remaining half of
the time, then
vav =
v1 + v2
.
2
(g) When a particle moves the first half of a distance
with a speed v1 and the second half of the distance with a speed v2, then
04_Kinematics 1_Part 1.indd 5
4.5
2v1v2
v1 + v2
(h) Similarly, when a particle covers one-third distance at speed v1, next one third with a speed v2
and the last one third at speed v3, then
vav =
3 v1v2 v3
v1v2 + v2 v3 + v3v1
Illustration 3
Calculate the average speed and the average velocity
in the following cases mentioned.
CASE-1: For a train that travels from one station to
another at a uniform speed of 40 kmh −1 and returns
to first station at a speed of 60 kmh −1 .
CASE-2: For a man who walks at a speed of 1 ms −1
for the first one minute and then runs at a speed of
3 ms −1 for the next one minute along a straight track.
CASE-3: For a man who walks 720 m at a uniform
speed of 2 ms −1 , then runs at a uniform speed of
4 ms −1 for 5 minute and then again walks at a speed
of 1 ms −1 for 3 minutes.
(Please consider all uniform speeds to be the average
speeds in respective intervals).
Solution
Average Speed =
Total Distance Travelled
Total Time Taken
( 2 ) ( 40 )( 60 )
2v1v2
= 48 kmh −1
=
v1 + v2
40 + 60
Average Velocity = vav = 0
{∵ train returns to its station}
CASE-1: vav =
CASE-2: vav =
v1 + v2 1 + 3
=
= 2 ms −1
2
2
CASE-3: vav =
s1 + s2 + s3
t1 + t2 + t3
where s1 = 720 m and
t1 =
s1
= 360 s = 6 minute
v1
s2 = ( 4 ) ( 5 )( 60 ) = 1200 m , t2 = 300 s
11/28/2019 7:02:05 PM
4.6 JEE Advanced Physics: Mechanics – I
s3 = ( 1 ) ( 3 )( 60 ) = 180 m , t3 = 180 s
⇒
vav =
720 + 1200 + 180 2100
=
360 + 300 + 180
840 ⇒
vav =
210 10
=
= 2.5 ms −1
84
4
⇒
Solution
Let us first draw a pictorial representation to the
problem.
2l
3
C
t1
v0
t2
B
D
t2
2
v1
t2
2
v2
For AC
l
= v0 t1
3
⇒
t1 =
l
…(1)
3v0
For CB
2l
= CD + DB
3
⇒
2l
⎛t ⎞
⎛t ⎞
= v1 ⎜ 2 ⎟ + v2 ⎜ 2 ⎟
⎝
⎠
⎝ 2⎠
3
2
⇒
t2 =
4l
…(2)
3 ( v1 + v2 )
Since, average velocity is defined as
l 2l
Total Displacement 3 + 3
vav =
=
Total Time
t1 + t2
04_Kinematics 1_Part 1.indd 6
4v0 + v1 + v2
The speed of a particle at a particular instant of time is
called the instantaneous speed. It is a scalar quantity.
A particle traversed one third the distance with a
velocity v0 . The remaining part of the distance was
covered with velocity v1 for half the time and with a
velocity v2 for the remaining half of time. Assuming
motion to be rectilinear, find the mean velocity of the
particle averaged over the whole time of motion.
A
3v0 ( v1 + v2 )
Instantaneous Speed and
Instantaneous Velocity
Illustration 4
l
3
vav =
Δx dx i
=
=x
Δt → 0 Δt
dt
vinstantaneous = vins = v = lim
The velocity of a particle at a particular instant of
time is called the instantaneous velocity, denoted by
v . So,
Δx dx i
vinstantaneous = vins = v = lim
=
=x
Δt → 0 Δt
dt
(a) Please note that, in Physics, dot on a physical
quantity indicates its derivative w.r.t. time e.g.
dp i dx i dv i
F=
=p, v=
=x, a=
=v
dt
dt
dt
dv d 2 x ii
=
=x
Also, a =
dt dt 2
i ii
So, a = v = x
(b) The magnitude of the instantaneous velocity
is always equal to the instantaneous speed i.e.,
dx dx
v= v =
≠
, because in general dx ≠ dx .
dt
dt
Acceleration ( a )
The rate of change of velocity with time is called
acceleration.
Δv v2 − v1
(a) Average Acceleration = aav =
=
Δt t2 − t1
(b) Instantaneous Acceleration is
Δv dv i
ains = a = lim
=
=v
Δt → 0 Δt
dt
dv d 2 x
So, a =
= 2
dt dt
i ii
⇒
a
=v=x
11/28/2019 7:02:09 PM
Chapter 4: Kinematics I
Conceptual Note(s)
(a) In 3-Dimensional space
If r = xi+ yj + zk
dr ⎛ dx ⎞
⎛ dy ⎞
⎛ dz ⎞
then v =
k⎜ ⎟
= i ⎜ ⎟ + j ⎜ ⎟ + ⎝
⎠
⎝
⎠
⎝ dt ⎠
dt
dt
dt
⇒ v = v x iˆ + v y ĵj + v z kˆ
d 2 r dv
and a = 2 =
dt
dt
⎛ d2 y ⎞
⎛ d2 z ⎞
⎛ d2 x ⎞
k⎜ 2 ⎟
⇒ a = i ⎜ 2 ⎟ + j ⎜ 2 ⎟ + ⎝ dt ⎠
⎝ dt ⎠
⎝ dt ⎠
⇒ a = a x iˆ + a y ĵj + a z kˆ
dv
is the magnitude of total acceleration. While
(b)
dt
d v
represents the time rate of change of speed
dt
(called the tangential acceleration, a component
of total acceleration) as v = v .
(c) These two are equal in case of one dimensional
motion (without change in direction).
(d) In case of uniform circular motion speed remains
constant while velocity changes.
d v
dv
Hence,
= 0 while
≠0
dt
dt
d v
(e)
≠ 0 implies that speed of particle is not
dt
constant. Velocity cannot remain constant if
­
dv
speed is changing. Hence,
cannot be zero
dt
dv
in this case. So, it is not possible to have
=0
dt
d v
while
≠ 0.
dt
Factors Affecting Acceleration
of a Body
The acceleration changes, when the velocity of the
body changes. The change in velocity may be due to
any of factors listed below.
(a) If the magnitude of velocity changes but direction remains the same.
04_Kinematics 1_Part 1.indd 7
4.7
(b) If the direction of the velocity changes but magnitude remains the same.
(c) If both magnitude and direction of the velocity
change.
Conceptual Note(s)
(a) A particle moving with uniform velocity has zero
acceleration i.e. it neither changes in magnitude
nor its direction.
(b) A particle moves with uniform acceleration if rate
of change of velocity is constant.
(c) A body is subjected to Retardation or Deceleration
when acceleration acts opposite to velocity.
(d) In case of a body subjected to retardation, we
take
(i) a as negative, if velocity is taken as positive.
(ii) a as positive, if velocity is taken as negative.
(e) In case of a body subjected to acceleration, we
take
(i) a as positive, if velocity is taken as positive.
(ii) a as negative, if velocity is taken as negative.
(f) However, a body at rest will be in accelerated
motion irrespective of the sign of acceleration.
(g) For a body moving in straight line with uniform
acceleration the average acceleration and instantaneous value of acceleration have same value.
(h) If a particle has an acceleration a1 for a time t1
and an acceleration a2 for a time t2, then average
a t +a t
acceleration is aav = 1 1 2 2
t1 + t2
Illustration 5
A body moves along a straight line. Its distance x
from a point on its path at a time t after passing that
point, is given by xt = 8t 2 − 3t 3 where xt is in meter
and t in second. Find
(a) the instantaneous velocity at t = 1 s
(b) instant and position at which the body is at rest
(c) the acceleration at t = 4 s
(d) the average velocity and average speed during
the interval t = 0 s to t = 4 s
Solution
⇒
x = 8t 2 − 3t 3 …(1)
dx
v=
= 16t − 9t 2 …(2)
dt
11/28/2019 7:02:13 PM
4.8 JEE Advanced Physics: Mechanics – I
a = 16 − 18t …(3)
(a) v t=1 s = 16 − 9 = 7 ms −1
(b) The body is at rest, when v = 0
⇒
16t − 9t 2 = 0 vav =
16
s
9
So, the particle is initially at rest. However at
t = 0 , a = −18 ms −2 which accelerates the particle to move.
⇒
So, from the table, we observe that the particle
16
s as shown.
reverses its direction of motion at t =
9
Hence average speed is
t=0, t=
2
⇒
–8.43 m
vav = 20.21 ms −1 Alternatively,
3
16
9
⇒ Average Velocity = −16 ms v = 0 at t = 0 (initially) and
16
s{point of reversal of motion}
9
Initial
Point
of Zero
Velocity
Final
Afterwards
t
0
16
s
9
4s
t>4
x = 8t2 - 3t3
0
8.43 m
-64 m
-
v = 16t - 9t2
0
0
-80 ms–1
-
a = 16 - 18t
16 ms–2
-16 ms–2
-56 ms–2
-
Nature of
Motion
Accelerating Accelerating Accelerating Accelerating
in Positive
in Negative in Negative
Direction
Direction
Direction
Average Speed =
= 20.21 ms −1
4
CHECK YOURSELF!
Illustration 6
A particle travels along a straight line with a v
­ elocity
v = ( 12 − 3t 2 ) ms −1 , where t is in seconds. When
t = 1 s , the particle is located 10 m to the left of the
origin. Calculate the
(a) acceleration when t = 4 s
(b) displacement from t = 0 to t = 10 s and
(c) distance the particle travels from t = 0 to t = 10 s.
Solution
(a) v = 12 − 3t 2 …(1)
⇒
a=
dv
= −6t
= −24 ms −2
dt
t= 4
(b) Further v =
x
⇒
dx
dt
t
t
∫ dx = ∫ v dt = ∫ ( 12 − 3t ) dt
2
−10
04_Kinematics 1_Part 1.indd 8
16
9
0
For calculating the average speed we must first
calculate the zeros of velocity i.e., the times at
which the particle is at rest (or momentarily at
rest i.e., at the point of reversal of motion).
In this problem, we observe that
v = 0 at t =
4
∫ v dt + ∫ v dt
−1
+
⎛ 16 s⎛
A ⎛t =
9
B
8( 4) − 3( 4) − 0
⇒ Average Velocity =
4
Positive
direction
O
C
t=4s
⇒
2
8.43 m
t=0
3
x
− x t=0
(d) Average Velocity = t = 4
4−0
4
–64 m
⎛ 16 ⎞
⎛ 16 ⎞
x = 0 and x = 8 ⎜ ⎟ − 3 ⎜ ⎟ = 8.43 m
⎝ 9 ⎠
⎝ 9 ⎠
(c) a t= 4 s = 16 − 18 ( 4 ) = −56 ms −2
vav =
Total Distance Covered
Total Time Taken
8.43 + − 8.43 + − 64
⎛
⇒
1
1
11/28/2019 7:02:19 PM
Chapter 4: Kinematics I
⇒
x + 10 = 12t − t 3 − 11
⇒
x = 12t − t 3 − 21
(c) From equation (1), we have
v = 0 when 12 − 3t 2 = 0
⇒
901 m
21 m
t=0
t = 10 s
–901
⇒
t= 0
= −21 m and x
5m
t= 10
–5 0
t=2s
So, x
t=2s
–21
So, x
4.9
3
t= 2
= 12 ( 2 ) − ( 2 ) − 21 = −5 m
⇒ Total Distance
x(m)
( xtotal ) = ( 21 − 5 ) + ( 901 − 5 ) = 912 m
= −901 m
Δx = −901 − ( −21 ) = −880 m
Test Your Concepts-I
Based on Displacement, velocity, Acceleration, Average speed and velocity
(Solutions on page H.65)
1. The acceleration of a particle as it moves along a
straight line is given by a = ( 2t − 1) ms −2 , where
t is in seconds. If x = 1 m and v = 2 ms −1 when
t = 0 , determine the particle’s velocity and position when t = 6 s. Also, determine the total distance the particle travels during this time period.
2. A particle moves in a straight line with a uniform
acceleration a. Initial velocity of the particle is zero.
Find the average velocity of the particle in first s
metre.
3. In one second a particle goes from point A to point
B moving in a semicircle (see figure). Find the magnitude of average velocity.
5.
6.
A
1m
7.
B
4. A particle is moving along a straight line
such that its position from a fixed point is
x = ( 12 − 15t 2 + 5t 3 ) m , where t is in seconds.
Determine the total distance travelled by the
04_Kinematics 1_Part 1.indd 9
8.
particle from t = 1 s to t = 3 s. Also, find the average speed of the particle during this time interval.
The position (x) of a particle, in metre, moving
along the x-axis depends on the time t, in seconds
as x = ct 2 − bt 3 , where c = 3 units and b = 2 units.
Calculate the
(a) units of c and b.
(b) time taken by the particle to reach its maximum positive x value.
(c) the distance travelled and the displacement of
the particle from t = 0 to t = 4 s.
(d) the velocity and acceleration at t = 0 , 1, 2, 3
and 4 second .
The position of a particle along a straight line is
given by x = ( t 3 − 9t 2 + 15t ) m , here t is in second.
Determine its maximum acceleration and maximum velocity during the time interval 0 ≤ t ≤ 10 s.
At time t = 0 , the position vector of a particle moving in the x -y plane is 5î m . At time t = 0.02 s ,
its position vector has becomeiˆ 5.1iˆ + 0.4 ˆj m .
Determine the magnitude of the average velocity
( vav ) during this interval and the angle q made by
the average velocity with the positive x-axis.
A particle travels along a straight line path such that
in 4 s it moves from an initial position x A = −8 m
to a position xB = +3 m . Then in another 5 s it
moves from xB to x C = −6 m . Determine the
11/28/2019 7:02:25 PM
4.10 JEE Advanced Physics: Mechanics – I
particle’s average velocity and average speed during the 9 s time interval.
9. A particle moves along a horizontal path such that
its velocity is given by v = ( 3t 2 − 6t ) ms −1 , where t
is the time in seconds. If it is initially located at the
­origin O, determine the
(a)distance travelled by the particle during the
time interval t = 0 to t = 3.5 s
(b)particle’s average velocity and average speed
during this time interval.
UNIFORMLY ACCELERATED
MOTION SYSTEMS
Consider a particle moving in a straight line with
constant acceleration a . If u is the initial velocity, v
is the velocity at time t , s is the displacement ( = Δx )
in time t and snth is the displacement in the nth second of motion, then equations governing the motion
of such a particle are
v = u + at
s = Δx = ut +
1 2
at
2
v 2 − u2 = 2 as
s=
u+v
t
2
snth = u +
1
a ( 2n − 1 )
2
For constant acceleration, a , the equations of motion
are written as
v = u + at
1
s = ut + at 2
2
v 2 − u2 = 2 a ⋅ s OR v ⋅ v − u ⋅ u = 2 a ⋅ s
⎛ u+v⎞
s=⎜
t
⎝ 2 ⎟⎠
where u = initial velocity of the particle
v = final velocity of the particle at time t
s = Δr is the displacement of the particle
04_Kinematics 1_Part 1.indd 10
10. The position of a particle along a straight line is
given by x = ( 1.5t 3 − 13.5t 2 + 22.5t ) m, where t is
in seconds. Determine the position of the particle
when t = 6 s and the total distance it travels during
the 6 s time interval.
11. The velocity of a particle moving in a straight
line decreases at the rate of 3 ms −1 per meter of
displacement at an instant when the velocity is
10 ms −1 . Calculate the acceleration of the particle
at this instant.
Conceptual Note(s)
(a) All the above equations of motion are to be
applied only when the motion is uniformly
accelerated.
(b) For applying the above equations greater care
has to be taken about the direction of the vector
quantities involved.
(c) Please note that this nth second has a duration of
1 second. So, snth is the distance travelled (or the
displacement) in 1 sec, hence
[ snth ] = LT
−1
Directions of Vectors in Straight
Line Motion
In straight line motion, all the vectors (position, displacement, velocity and acceleration) will have only
one component (along the line of motion) and there
will be only two possible directions for each vector.
For example
(a) If a particle is moving in a horizontal line along
x-axis, then, the two directions are right and left.
Any vector directed towards right can be represented by a positive number and towards left can
be represented by a negative number.
–
+
Line of motion
11/28/2019 7:02:29 PM
Chapter 4: Kinematics I
(b) For vertical or inclined motion, upward direction can be taken as positive and downward as
negative.
Line of motion
For AC
⇒
n
io
m
ot
of
–
⎛ l⎞
2
vm
− u2 = 2a ⎜ ⎟
⎝ 2⎠
2
vm
− u2 = al …(1)
For CB
⇒
Li
–
+
ne
+
4.11
For objects moving vertically near the surface
of the earth, the only force acting on the particle is its weight ( mg ) i.e., the gravitational pull
of the earth. Hence acceleration for this type of
motion will always be a = − g i.e., a = −9.8 ms −2
(­negative sign, because the force and acceleration
are directed downwards. If we select upward
direction as positive).
⎛ l⎞
2
v 2 − vm
= 2a ⎜ ⎟
⎝ 2⎠
2
v 2 − vm
= al …(2)
Equating (1) and (2), we get
vm =
u2 + v 2
…(3)
2
(b) Since tA→C = 2tC →B
Let tC →B = t , then tA→C = 2t
⇒
l ⎛ u + vm ⎞
=⎜
⎟ 2t {for AC }…(4)
2 ⎝ 2 ⎠
⇒
l ⎛ vm + v ⎞
=⎜
⎟ t 2 ⎝ 2 ⎠
(a) If acceleration is in same direction as velocity,
then speed of the particle increases.
(b) If acceleration is in opposite direction to the velocity then speed decreases i.e., the particle slows
down. This situation is known as retardation.
⇒
u + vm =
⇒
⇒
v m = v − 2u
⇒
2
vm
= v 2 + 4u2 − 4uv
Illustration 7
Since, vm =
A point moving with constant acceleration from A to
B in a straight line AB has velocities u and v at A
and B respectively.
⇒
Conceptual Note(s)
{for CB }…(5)
vm + v
2
2u + 2vm = vm + v
u2 + v 2
2
u2 + v 2
= v 2 + 4u2 − 4uv
2
(a) Find its velocity at C , the midpoint of A and B .
v
(b) Find the ratio , if time taken from A to C is
u
twice the time to go from C to B.
⇒
⇒
u2 + v 2 = 2v 2 + 8u2 − 8uv
v 2 − 7 uv − uv + 7 u2 = 0
Solution
⇒
⇒
⇒ ( v − u ) ( v − 7 u ) = 0
As motion is an accelerated motion
So, v − u ≠ 0
⇒ v = 7 u
(a) Let vm be the velocity at C , the midpoint of A
and B .
A
04_Kinematics 1_Part 1.indd 11
l
2
u
l
2
C
vm
B
v
⇒
v 2 − 8uv + 7 u2 = 0
v ( v − 7u ) − u ( v − 7u ) = 0
v
=7
u
11/28/2019 7:02:38 PM
4.12 JEE Advanced Physics: Mechanics – I
Illustration 8
+
−1
A body starts with an initial velocity of 10 ms and
moves along a straight line path with constant acceleration. When the velocity of the body is 50 ms −1 the
acceleration is reversed in direction. Find the velocity
of the particle as it reaches the starting point.
P
B
l
For P
al = 1200 …(1)
Now, when the particle just attains the velocity of
50 ms −1 at A , acceleration is reversed in direction
i.e., it starts acting as retardation for 50 ms −1 which
will first be reduced to zero after travelling a distance
AB and then the particle will again be accelerated
from B to A with the same acceleration. So, velocity of the particle again at A will be 50 ms −1 , but in
opposite direction.
a
10 ms–1
l
a
A
50 ms–1
B
v=0
−
⇒
1
2
( u2 − u1 ) t + ( a2 − a1 ) t 2 = 0
⇒
⇒
v 2 = 4900 v = 70 ms −1 ⎛ u − u1 ⎞
t = 2⎜ 2
…(3)
⎝ a1 − a2 ⎟⎠
Now, substitute this value of t from (3) in any of
equations (1) or (2), we get
2
v 2 − 2500 = 2 ( 1200 ) l
1
= u2 t + a2 t 2 …(2)
2
2
l
⎡ ⎛ u − u1 ⎞ ⎤ 1 ⎡ ⎛ u2 − u1 ⎞ ⎤
= u1 ⎢ 2 ⎜ 2
⎟⎥
⎟ ⎥ + a1 ⎢ 2 ⎜
2
⎣ ⎝ a1 − a2 ⎠ ⎦ 2 ⎣ ⎝ a1 − a2 ⎠ ⎦
v 2 − ( 50 ) = 2 al
⇒
l
1
= −u2 t + ( − a2 ) t 2
2
2
From (1) and (2), we get (by subtracting)
From A to O
⇒
{in the opposite direction}
So, v = 70 ms −1 from A to O
⇒
⇒
Illustration 9
Two particles P and Q move in a straight line AB
towards each other. P starts from A with a velocity u1 and an acceleration a1 . Q starts from B
with velocity u2 and an acceleration a2 . They pass
from each other at midpoint of AB and arrive at
other ends of AB with equal velocities. Prove that
( u1 + u2 ) ( a1 − a2 ) = 8 ( a1u2 − a2 u1 ) .
Let the particle P reach C in time t , then the particle
Q must reach C also in time t .
l
⎛ u − u1 ⎞ ⎡
⎛ u − u1 ⎞ ⎤
u1 + a1 ⎜ 2
= 2⎜ 2
2
⎝ a1 − a2 ⎟⎠ ⎢⎣
⎝ a1 − a2 ⎟⎠ ⎥⎦
l=
4 ( u2 − u1 )
2
( u1a1 − u1a2 + a1u2 − a1u1 )
( a1 − a2 )2
4 ( u2 − u1 )
l=
( a1u2 − a2 u1 ) …(4)
( a1 − a2 )2
Again, after reading the question carefully, we
observe that P reaches B and Q reaches A with
equal velocities, say v.
Then, for P , we have
v 2 − u12 = 2 a1l …(5)
For Q , we have
v 2 − u22 = 2 a2 l …(6)
Solution
04_Kinematics 1_Part 1.indd 12
Q
For Q
( 50 )2 − ( 10 )2 = 2al
⇒
u2
l
1
= u1t + a1t 2 …(1)
2
2
For OA ( = l )
O
C
A
Solution
⇒
u1
⇒
u22 − u12 = 2 ( a1 − a2 ) l
11/28/2019 7:02:47 PM
4.13
Chapter 4: Kinematics I
Substituting value of l, from (4), we get
u22 − u12 = 2 ( a1 − a2 )
⇒
4 ( u2 − u1 )
( a1u2 − a2 u1 )
( a1 − a2 )2
( u1 + u2 ) ( a1 − a2 ) = 8 ( a1u2 − a2 u1 ) Illustration 10
A man is standing 40 m behind the bus. Bus starts
with 1 ms −2 constant acceleration and also at the
same instant the man starts moving with constant
speed 9 ms −1 . Find the time taken by man to catch
the bus.
speed during the interval is 72 kmh −1 . Calculate the
time for which the car was in uniform motion.
Solution
Since the car accelerates from zero to v and decelerates from v to zero at the same rate of 5 ms −2, so
it must travel equal distances during the accelerated
and the decelerated interval in equal times.
If t be the total time of journey, then
t = 25 = t1 + t2 + t1
⇒
25 = 2t1 + t2 …(1)
v
1 ms–2
A
vmax = v
x=0
t=0
40 m
t=0
x = 40
l1
Solution
x = 40 +
t2
…(1)
2
For man
x = 9t …(2)
From (1) and (2), we get
40 +
⇒
⇒
t2
= 9t
2
t − 18t + 80 = 0 t1
t = 10 s Illustration 11
A car starts moving rectilinearly, first with an acceleration of 5 ms −2 (initial velocity zero), then moves
uniformly and finally decelerating at the same rate till
it stops. The total time of journey is 25 s. The a­ verage
04_Kinematics 1_Part 1.indd 13
t
t1
vav = 72 kmh −1 = 20 ms −1 Total Distance Travelled l1 + l2 + l1
=
Total Time Taken
25
⇒
vav =
⇒
2l1 + l2 = ( 20 )( 25 ) = 500 ⇒
2l1 + l2 = 500 …(2)
where l1 =
5 2
t1 …(3)
2
Also, for OA , we have
For AB, we have
t − 10t − 8t + 80 = 0 OR
t2
C
Furthermore, the average speed
2
t=8s
l1
N
v = 5t1 …(4)
2
⇒ ( t − 8 ) ( t − 10 ) = 0 ⇒
l2
M
O
Let the man catches the bus at time t .
For bus
1
x = x0 + ut + at 2
2
1
⇒ x = 40 + 0 ( t ) + ( 1 ) t 2
2
⇒
B
l2 = vt2 = 5t1t2 …(5)
⇒
⎛5 ⎞
2 ⎜ t12 ⎟ + 5t1t2 = 500 ⎝2 ⎠
⇒
5t12 + 5t1 ( 25 − 2t1 ) = 500
⇒
t12 + t1
⇒
t12 − 2t12 + 25t1 = 100 {substituting in (2)}
( 25 − 2t1 ) = 100 11/28/2019 7:02:54 PM
4.14 JEE Advanced Physics: Mechanics – I
⇒
t12 − 25t1 + 100 = 0 ⇒
t12 − 20t1 − 5t1 + 100 = 0
⇒
t1 ( t1 − 20 ) − 5 ( t1 − 20 ) = 0
⇒
( t1 − 5 ) ( t1 − 20 ) = 0 ⇒
t1 = 5 s
OR
t1 = 20 s
t1 = 5 s
⇒
t=
Hence the car was in uniform motion for 15 s .
Illustration 12
A police inspector in a car is chasing a pickpocket
an a straight road. The car is going at its maximum
speed v (assumed uniform). The pickpocket rides on
the motorcycle of a waiting friend when the car is at
a distance d away and the motorcycle starts with a
constant acceleration a . Show that the pick pocket
will be caught if v ≥ 2 ad .
Solution
Suppose the pickpocket is caught at a time t after
motorcycle starts. The distance travelled by the
motorcycle during this interval is
1 2
at …(1)
2
During this interval the car travels a distance
s + d = vt …(2)
v
v 2 ≥ 2 ad
⇒
v ≥ 2 ad When a situation demands our immediate action, it
takes some time before we really respond. Reaction
time is the time a person takes to observe, think
and act.
Illustration 13
A driver takes 0.20 s to apply the brakes after he sees
a need for it. This is called the reaction time of the
driver. If he is driving car at a speed of 5.4 kmh −1
and the brakes cause a deceleration of 6 ms −2 , find
the distance travelled by the car after he sees the need
to put the brakes.
Solution
During the reaction time of 0.20 s , the car continues
to move with a speed of 54 kmh −1 i.e. 15 ms −1 .
So, distance travelled by car during this time is
s1 = ( 15 )( 0.2 ) = 3 m
Now, when brakes are applied, the distance s2
t­ ravelled by the car is
2
0 2 − ( 15 ) = 2 ( −6 ) s2
a
v=0
d
⇒
1 2
at + d = vt
2
s2 =
225
m = 18.75 m
12
So, total distance covered by the car is
From (1) and (2), we get
04_Kinematics 1_Part 1.indd 14
v ± v 2 − 2 ad
a
REACTION TIME
t2 = 15 s s=
at 2 − 2vt + 2d = 0 The pickpocket will be caught if t is real and positive.
This will be possible if
But if t1 = 20 s , then total time becomes greater than
25 s which is impossible. So,
⇒
⇒
s = s1 + s2 = 3 + 18.75
⇒
s = 21.75 m 11/28/2019 7:03:00 PM
Chapter 4: Kinematics I
4.15
Test Your Concepts-II
Based on Constant Acceleration
1. A particle moving in a straight line with constant
acceleration travels a distance x, y and z during
the pth, qth and rth second respectively. Prove
that
(q − r )x + (r − p )y + ( p − q )z = 0
2. A car accelerates from rest at a constant rate a for
some time, after which it decelerates at a constant
rate b, to come to rest. If the total time elapsed is
t seconds, evaluate
(a) the maximum velocity reached and
(b) the total distance travelled.
3. A particle starts moving from the position of rest
under a constant acceleration. If it travels a distance x in t sec, what distance will it travel in next
t sec?
4. In a car race, car A takes time t less than car B and
passes the finishing point with a velocity v more
than the velocity with which car B passes the point.
Assuming that the cars start from rest and travel
with constant accelerations a1 and a2, show that
v = t a1a2 .
5. Two cars start off to race with velocities v1 and v2
and travel in a straight line with uniform accelerations a1 and a2. If the race ends in a dead heat,
prove that the length of the course is
2 ( v1 − v2 ) ( v1a2 − v2 a1 )
(Solutions on page H.67)
8. To stop a car, first you require a certain reaction
time to begin braking and then the car slows
under the constant braking deceleration. Suppose
that the total distance moved by your car during
these two phases is 56.7 m when its initial speed
is 80.5 kmh−1 and 24.4 m when its initial speed is
48.3 kmh−1 . What is
(a) your reaction time and
(b) the magnitude of the deceleration?
9. A particle, starting from rest moves in a straight line
with constant acceleration. After time t0 the acceleration changes its direction without any change
in its magnitude. Determine the time t from the
beginning of motion in which the particle returns
to the initial position.
10. The velocity v of a particle moving in a
straight line varies with its displacement x as
v = ( 4 + 4 x ) ms −1. Displacement of particle at
time t = 0 is x = 0 . Find displacement of particle
at time t = 2 s.
11. A particle starts from rest and traverses a distance
l with uniform acceleration, then moves uniformly
over a further distance 2l and finally it comes to
rest after moving a further distance 3l under uniform retardation. Assuming the entire path to be a
straight line find the ratio of the average speed over
the journey to the maximum speed on its way.
v
( a1 − a2 )2
6. A car moving with constant acceleration covered
the distance between two points 60 m apart in 6 s.
Its speed as it passes the second point was 15 ms −1.
(a) What was the speed at the first point?
(b) What was the acceleration?
(c) At what prior distance from the first was the
car at rest?
7. A train stopping at two stations 4 km apart takes
4 minute on the journey from one station to the
other. Assuming that it first accelerates with a
uniform acceleration x and then that of uniform
1 1
retardation y, prove that + = 2.
x y
04_Kinematics 1_Part 1.indd 15
A
vmax
l
O
B
2l
M
t1
3l
N
t2
C
t
t3
12. An a particle travels along the inside of straight
hollow tube, 2 m long, of a particle accelerator.
Under uniform acceleration, how long is the particle in the tube if it enters at a speed of 1000 ms −1
and leaves at 9000 ms −1 . What is its acceleration
during this interval?
11/28/2019 7:03:03 PM
4.16 JEE Advanced Physics: Mechanics – I
13. A sports car passing a police check-point at
60 kmh−1 immediately started slowing down
uniformly until its speed was 40 kmh−1 . It continued to move at this speed until it was passed by a
police car 1 km from the check-point. This police
car had started from rest at the check-point at
the same instant as the sports car had passed the
check-point. The police car had moved with constant acceleration until it had passed the sports car.
Assuming that the time taken by the sports car in
slowing down from 60 kmh−1 to 40 kmh−1 was
equal to the time that it travelled at constant speed
before passed by the police car, find
(a)the time taken by the police car to reach the
sports car,
(b)the speed of the police car at the instant when
it passed the sports car,
(c)the time measured from the check-point when
the speeds of the two cars were equal.
14. Two railway stations A and B are 50 km apart and
are served by electric trains which can decelerate at
5 kmh−1 per second, and accelerate 3 kmh−1 per
second. The maximum speed is 90 kmh−1 . There
are twelve intermediate stations all more than a
km apart. Find the least time which can be taken to
made the journey from A to B
(a) by a fast non-stop train and
1
(b)by a slow train which stops minute at every
2
station.
15. A truck starts from rest with an acceleration of
1.5 ms −2 while a car 150 m behind starts from rest
with an acceleration of 2 ms −2 . How long will it
take before both the truck and car side by side, and
how much distance is travelled by each?
16. Two cars travelling towards each other on a straight
road at velocity 10 ms −1 and 12 ms −1 respectively.
When they are 150 m apart, both drivers apply
their brakes and each car decelerates at 2 ms −2
until it stops. How far apart will they be when they
have both come to a stop?
17. Two motor cars start from A simultaneously and
reach B after 2 hours. The first car travelled half the
distance at a speed of 30 kmh−1 and the other half
at a speed of 60 kmh−1 . The second car covered
the entire distance with a constant acceleration.
At what instant of time, the speeds of both the
04_Kinematics 1_Part 1.indd 16
vehicles is the same? Will one of them overtake the
other enroute?
18. A point travelling along a straight line traversed one
third the distance with a velocity v0. The remaining
part of the distance was covered with velocity v1
for half the time and with velocity v2 for the other
half of the time. Find the mean velocity of the point
averaged over the whole time of motion.
19. A man runs at a speed of 4 ms −1 to overtake a
standing bus. When he is 6 m behind the door
( at t = 0 ) , the bus moves forward and continues
with a constant acceleration of 1.2 ms −2 . How long
does it take for the man to go to the door? If he
was initially 10 m behind the door, can he catch the
bus?
20. Two cars A and B start off to a race on a straight
path with initial velocities 8 ms −1 and 5 ms −1
respectively. Car A moves with uniform acceleration of 1 ms −2 and B moves with uniform acceleration of 1.1 ms −2 . If both the cars reach the winning
post together, find the length of the track. Also find
which of the two cars was ahead 10 s before the
finish.
21. A car starts from rest and moves with a constant
acceleration of 1.5 ms −2 until it achieves a velocity
of 25 ms −1 . It then travels with constant velocity for
60 seconds . Determine the average speed and the
total distance travelled.
22. A car is to be hoisted by elevator to the fourth
floor of a parking garage, which is 14 m above the
ground. If the elevator can accelerate at 0.2 ms −2 ,
decelerate at 0.1 ms −2 and reach maximum speed
of 2.5 ms −1 , determine the shortest time to make
the lift, starting from rest and ending at rest.
23. The driver of a car wishes to pass a truck that is
travelling at a constant speed of 20 ms −1 . Initially,
the car is also travelling at 20 ms −1 . Initially, the
vehicles are separated by 25 m, and the car pulls
back into the truck’s lane after it is 25 m ahead of
the truck. The car is 5 m long, and the truck is 20 m
long. The car’s acceleration is a constant 0.6 ms −2 .
(a)How much time is required for the car to pass
the truck?
(b)What distance does the car travel during this
time?
(c) What is the final speed of the car?
11/28/2019 7:03:08 PM
Chapter 4: Kinematics I
24. A train of length l = 348 m starts moving rectilinearly with constant acceleration a = 3 cms −2 . After
30 second a ball is dropped by the driver (event 1)
and after 60 second another ball is dropped by the
guard (event 2). How and at what constant velocity
v should a driver drive his car parallel to the train
so that he observes both the events to occur at the
same point. Neglect the length of the car.
EQuations of Motion for Variable
Acceleration
CASE-1: When acceleration a of the particle is a
function of time
Since acceleration of a particle is a function of time,
i.e., a = f ( t )
⇒
dv
= f (t ) dt
⇒
dv = f ( t ) dt {by definition}
Integrating both sides within suitable limits, we have
v
t
u
0
∫ dv = ∫ f ( t ) dt
t
⇒
v = u+
∫ f ( t ) dt
0
⇒
{by definition}
Multiply and divide the L.H.S. by dx and rearrange,
we get
dv dx
⋅
= f (x)
dx dt
Integrating both sides within suitable limits, we have
∫
x
vdv =
u
04_Kinematics 1_Part 1.indd 17
∫
x0
⇒
v 2 u2
−
=
2
2
f ( x ) dx
x
∫ f ( x ) dx
x0
x
⇒
v 2 = u2 + 2
∫ f ( x ) dx
x0
CASE-3: When acceleration a of the particle is a
function of velocity
Since the particle is a function of velocity, i.e.,
a = f (v)
dv
= f (v) {by definition}
dt
dv
⇒ dt =
f (v)
Integrating both sides within suitable limits, we have
⇒
∫
dv
= f (x) dt
v
25. A particle moves rectilinearly with constant acceleration. The displacement, measured from a convenient fixed position, is 2 m at time t = 0 and
is zero when t = 10 s. If the particle reverses its
direction of motion at t = 6 s, determine the acceleration a and the velocity v when t = 10 s.
t
CASE-2: When acceleration a of the particle is a
­function of distance
Since acceleration of a particle is a function of
­distance, i.e., a = f ( x )
4.17
v
dt =
0
∫
u
v
⇒
t=
dv
f (v)
dv
∫ f (v)
u
In this case we shall get v as a function of time i.e.,
v(t )
Otherwise
v
dv
= f (v)
dx
Rearranging and integrating both sides within suitable limits, we have
x
∫
x0
v
dx =
vdv
∫ f (v)
u
11/28/2019 7:03:12 PM
4.18 JEE Advanced Physics: Mechanics – I
v
⇒
x − x0 =
∫
u
v
⇒
x = x0 +
vdv
f (v)
Illustration 15
vdv
∫ f (v)
In this case we shall get x as a function of velocity
i.e., x ( v )
u
Illustration 14
An object starts from rest at t = 0 and accelerates at a
rate given by a = 6t . What is
(a) its velocity and
(b) its displacement at any time t ?
For a particle moving along x-axis, velocity is given
as a function of time as v = 2t 2 + sin t. At t = 0 , particle is at origin. Find the position and acceleration
as a function of time. Also calculate initial speed and
acceleration of the particle.
Solution
⇒
v = 2t 2 + sin t
dx
= 2t 2 + sin t
dt
x
⇒
t
∫ dx = ∫ ( 2t + sin t ) dt
2
0
0
2
x = t 3 − cos t + 1
3
Solution
⇒
As acceleration as a function of time is
Since v = 2t 2 + sin t
a = 6t
⇒
⇒
dv
= 6t
dt
⇒
dv = 6tdt v
⇒
2
∫ dv = 6∫ t dt
0
v=6
⇒
v = 3t 2 Now v =
Initial Acceleration is a t= 0 = 4 ( 0 ) + cos ( 0 ) = 1 ms −2
Illustration 16
A particle is subjected to an acceleration a = α t + βt 2 ,
where α and β are positive constants. The position
and velocity of the particle at t = 0 are x0 and v0
respectively. Find the expression for the velocity ( v )
and position ( x ) of the particle at time t .
t2
= 3t 2
2
⇒
dv
= 4t + cos t
dt
So, initial speed is v t= 0 = u = 2 ( 0 ) + sin ( 0 ) = 0 and
t
0
a=
dx
dt
Solution
Since a = α t + βt 2 …(1)
⇒
dx
= 3t 2
dt
⇒
⇒
dx = 3t 2 dt ⇒
x2
t
Integrating, we get
x1
0
⇒
⇒
⇒
⇒
∫ dx = 3∫ t dt
⇒
3 t
3t
3
x2 − x1 = Δx = t 3 Δx = t 04_Kinematics 1_Part 1.indd 18
dv
= α t + βt 2
dt
dv = α tdt + βt 2 dt v
x2 − x1 = Δx =
3
0
t
t
∫ dv = α ∫ tdt + β∫ t dt
2
v0
0
⇒
v − v0 = α
⇒
v = v0 +
0
2
3
t
t
+β
2
3 α t 2 βt 3
+
…(2)
2
3
11/28/2019 7:03:21 PM
Chapter 4: Kinematics I
Since v =
Illustration 18
dx
dt
2
For a particle moving along x-axis, acceleration is
given as a = 2v 2 . If the speed of the particle is v0 at
x = 0 , find speed as a function of x .
3
dx
αt
βt
= v0 +
+
dt
2
3 2
αt
βt 3
⇒ dx = v0 dt +
dt +
dt
2
3
Integrating, we get
⇒
x
⇒
4.19
t
α
t
β
Solution
a = 2v 2
⇒
t
∫ dx = v ∫ dt + 2 ∫ t dt + 3 ∫ t dt
0
0
x0
2
0
3
0
⇒
dv
= 2v 2
dt
dv dx
×
= 2v 2
dx dt
⇒
x − x0 = v0 t +
α t 3 βt 4
+
6
12 ⇒
v
⇒
x = x0 + v0 t +
α t 3 βt 4
+
…(3)
6
12
⇒
dv
= 2v
dx
So, equations (2) and (3) give us the required expression for v and x .
v
⇒
Illustration 17
For a particle moving along x-axis, acceleration is
given as a = x . Find the position as a function of time
if at t = 0 , x = 1 m and v = 1 ms −1
Solution
∫
v0
dv
= 2v 2
dx
dv
=
v
x
∫ 2dx
0
x
v
log e v v = ( 2x ) 0
0
⇒
⎛ v ⎞
log e ⎜ ⎟ = 2x
⎝ v0 ⎠
⇒
v = v0 e 2 x {∵ If log e x = α ⇒ x = eα }
a=x
Since, a =
⇒
⇒
dv
dv
=v
dt
dx
v
dv
=x
dx
v
x
∫
v dv =
0
⇒
⇒
⇒
∫
dx
=
x
t
∫ dt
0
x
log e x 1 = t 1
⇒
0
⇒
log e x = t ⇒
x = et 04_Kinematics 1_Part 1.indd 19
For a particle moving along x-axis, acceleration is
given as a = v . Find the position as a function of time
if at t = 0 , x = 0 and v = 1 ms −1 .
Solution
x dx
v2 x 2
=
2
2 v=x
dx
=x
dt
x
⇒
∫
Illustration 19
⇒
a=v
dv
=v
dt
v
⇒
∫
1
Since,
dv
=
v
∫
t
∫ dt
dv
= log e v
v
0
v
t
⇒
log e v 1 = t 0
⇒
⎛ v⎞
log e ⎜ ⎟ = t
⎝ 1⎠
v = et ⇒
11/28/2019 7:03:31 PM
4.20 JEE Advanced Physics: Mechanics – I
dx
= et
dt
⇒
⇒
t
dx = e dt x
⇒
6 − 3v = 6 e −3t
⇒
v = 2(1 − e −3t ) ms −1
(d) For acceleration to be half the initial value i.e.,
t
∫ dx = ∫ e dt
⇒
⇒
6
= 3 ms −2
2
3 = 6 − 3v(t)
⇒
v(t) = 1 ms −1
t
0
0
⇒
x = et − e0 ⇒
x = et − 1 a=
Illustration 21
Illustration 20
The motion of a body is given by the equation
dv(t)
= 6 − 3v(t) , when v ( t ) is the speed in ms −1 and
dt
t in seconds. If the body was at rest at t = 0 ; Then
test the correctness of the following result
(a) the terminal speed is 2 ms −1
(b) the magnitudes of initial acceleration is 6 ms −2
(c) the speed varies with time as v(t) = 2(1 − e −3t ) ms −1
(d) the speed is 1 ms −1 when the acceleration is half
the initial value
Solution
(a) the time dependence of the velocity and the
acceleration of the particle.
(b) the mean velocity of the particle averaged over
the time that the particle takes to cover first s
meters of the path.
Solution
METHOD I
(a) v = α x
(a) The terminal speed is the speed at which acceleration is zero i.e.,
dv(t)
=0
dt
−1
⇒ v = 2 ms
(b) Since initially body is at rest hence
v(t ) = v(0 ) = 0
⇒
dv
= 6 ms −2
dt t= 0
v
dv
t
∫ 6 − 3v = ∫ dt
0
dv d
= (α x )
dt dt
α dx
α
=
v
2 x dt 2 x
{
dx
=v=α x
dt
⇒
a=
⇒
α2
⎛ α ⎞(
)
α
a=⎜
x
=
= constant
⎝ 2 x ⎟⎠
2
a=
dv
dt
⇒
a=
dv α 2
=
dt
2
⇒
dv =
v
0
v
⇒
1
t
− log e |6 − 3v| = t 0
3
0
⇒
⎛ 6 − 3v ⎞
= −3t
log e ⎜
⎝ 6 ⎟⎠
04_Kinematics 1_Part 1.indd 20
Since a =
∵
}
Further, we have
dv
= dt
(c)
6 − 3v
Integrating both sides
The velocity of a particle moving in the positive direction of x-axis varies as v = α x , where α is a positive
constant. Assuming that at moment t = 0 , the particle was located at the point x = 0 . Find
⇒
∫
0
⇒
α2
dt
2
α2
dv =
2
t
∫ dt
0
⎛ α2 ⎞
v=⎜
t
⎝ 2 ⎟⎠
11/28/2019 7:03:41 PM
Chapter 4: Kinematics I
⇒
s
s 2 s
vav = v = =
t
α
⇒
vav =
t
∫
v dt
(b) Since v = 0t
=
∫ dt
Total Distance Travelled s
=
Total Time Taken
t
0
To cover a distance s in time t with constant
α2
acceleration a =
, we have
2
s=
1 ⎛ α2 ⎞ 2
⎜
⎟t
2⎝ 2 ⎠
⇒
t=
2 s
α ⇒
v = vav =
A particle having a velocity v = v0 at t = 0 is
decelerated at the rate a = α v , where α is a
positive constant. After what time and distance will
the particle will be at rest?
Solution
s
⎛2 s⎞
⎜⎝
⎟
α ⎠
=
v = v0 at t = 0 and a = −α v
α s
2
Time taken by particle to come to rest:
Since v = α x , so we get
v2 = α 2 x
a=
0
⇒
we get
α2
2
Hence the motion is uniformly accelerated with zero
initial velocity ( u = 0 ) and c­onstant acceleration
⎛
α2 ⎞
⎜⎝ a =
⎟.
2 ⎠
⇒
⇒
⇒
t
0
−
1 v −1 2 + 1
α −1 2 + 1
t=−
t=−
0
t
= t0
0
1 ⎛ v1 2 ⎞
α ⎜⎝ 1 2 ⎟⎠
v0
v0
1
2
0−
=
v0
α
α
12
Distance travelled before coming to rest:
dv
= −α v
ds
vdv
−
= ds
α v
a=v
(a) Since v = at
⇒
v=
α 2t
2
⇒
a=
α2
= constant
2
⇒
0
⇒
2
1 2 1α 2
at =
t
2
2 2
2 s
t=
= time taken to cover first s metre
α
04_Kinematics 1_Part 1.indd 21
dv
∫ − α v = ∫ dt
v 2 = u2 + 2 as
⇒
dv
= −α v
dt
v0
Comparing this equation with the equation of motion
i.e.,
(b) s =
α s
2
Illustration 22
METHOD II
u = 0 and a =
4.21
1
−
vdv =
α
∫
v0
⇒
−
1 ⎛ v3 2 ⎞
α ⎜⎝ 3 2 ⎟⎠
0
∫ ds
S
0
0
= ss
v0
11/28/2019 7:03:48 PM
4.22 JEE Advanced Physics: Mechanics – I
1⎡
2
32⎤
0 − v0
⎥⎦ = 0 − s
α ⎢⎣
3
⇒
−
⇒
s=
2 32
v
3α 0 Illustration 23
The velocity of a particle travelling along a straight
line is v = v0 − kx , where k is constant. If x = 0 when
t = 0 , determine the position and acceleration of the
particle as a function of time.
Solution
Since v =
dx
dt
t
x
⇒
∫
dt =
0
0
t
dx
∫ v − kx
0
⇒
t=
⇒
v0
e =
v0 − kx
⇒
0
1 ⎛ v0 ⎞
ln
k ⎜⎝ v0 − kx ⎟⎠
kt
v0
−
x = ( 1 − e kt )
k
Now, v =
⇒
1
= − ln ( v0 − kx )
k
0
t
dx d ⎡ v0 (
⎤
=
1 − e − kt ) ⎥
dt dt ⎢⎣ k
⎦
v = v0 e − kt
Since, we know that
a=
⇒
(
dv d
=
v0 e − kt
dt dt
a = − kv0 e − kt
04_Kinematics 1_Part 1.indd 22
(a) For a particle having zero initial velocity if v ∝ t ,
s ∝ t 2 and v 2 ∝ s then acceleration of particle
must be constant i.e., particle is moving rectilinearly with uniform acceleration.
(b) For a particle having zero initial velocity if s ∝ t α ,
where α > 2, then particle’s acceleration increases
with time.
(c) For a particle having zero initial velocity if s ∝ t α ,
where α < 0, then particle’s acceleration decreases
with time.
Problem Solving Technique(s)
x
⇒
Conceptual Note(s)
)
When acceleration of particle is not constant, we go
for basic equations of velocity and acceleration, i.e.,
dx
dr
(a) v =
or sometimes v =
dt
dt
dv
(b) a =
dt
(c) dx = dr = vdt
(d) dv = adt
For one dimensional motion, above relations can be
written as under.
dx
(a) v =
dt
dv
dv
(b) a =
=v
dt
dx
(c) dx = vdt
(d) dv = adt or vdv = adx
11/28/2019 7:03:54 PM
Chapter 4: Kinematics I
4.23
Test Your Concepts-III
Based on variable Acceleration
1. A particle starts from rest and travels along a straight
v⎞
⎛
line with an acceleration a = ⎜ 30 − ⎟ ms −2 where
⎝
5⎠
v is in ms −1 . Determine the time when the velocity
2.
3.
4.
5.
6.
7.
of the particle is v = 30 ms −1 .
The acceleration (a) of a particle moving in a
straight line varies with its displacement (s) as
a = 2s. The velocity of the particle is zero at zero
displacement. Find the corresponding velocitydisplacement equation.
The velocity of a particle travelling along a straight
line is v = v0 − kx , where k is constant. Initially, it
is observed that the particle is at the origin of the
coordinate system, then determine the position
and acceleration of the particle as a function of
time.
The retardation of a particle moving in a straight
line is proportional to its displacement (proportionality constant being 4). Find the total distance
covered by the particle till it comes to rest. Given
that velocity of particle is v0 at zero displacement.
The acceleration (a) of a particle travelling
along a straight line varies with distance x as
a = ( 8 − 2 x ) ms −2 , where x is in meters. Assuming
that the particle starts from rest from the origin of
the coordinate system, calculate the velocity of the
particle at x = 2 m and the position of the particle
when the velocity is maximum. Also calculate the
maximum velocity.
The acceleration of a particle travelling along a
k
straight line is a = , where k is a constant. If x = 0 ,
v
v = v0 initially, determine the velocity of the
particle as a function of time t.
The motion of a body falling from rest in a resisting medium is described by the equation given as
dv
= A − Bv ; where A and B are constants. Find the
dt
initial acceleration and the velocity at which the
acceleration is zero. Find the velocity at any instant
of time.
04_Kinematics 1_Part 1.indd 23
(Solutions on page H.73)
8. A particle is moving along a straight line such that
its speed is defined as v = ( −4 x 2 ) ms −1 , where x
is in meters. If x = 2 m when t = 0 , determine the
velocity and acceleration as functions of time.
9. The acceleration of a particle is given by
a ( t ) = ( 3 − 2t ) .
(a) Find the initial speed v0 such that the particle
will have the same x-coordinate at t = 5 s as it
had at t = 0 .
(b) What will be the velocity at t = 5 s ?
10. The acceleration of a particle travelling along a
straight line is a = ( 5e t ) ms −2 , where t is in seconds. If v = 0 , x = 0 when t = 0 , determine the
velocity and displacement of the particle as a function of t.
11. A particle moves along a straight line with a velocity v = ( 200 x ) mms −1 , where x is in millimeters.
Determine the acceleration of the particle at
x = 2000 mm . How long does the particle take to
reach this position if x = 500 mm when t = 0 ?
12. A particle has acceleration varying with time t as
a = 2tiˆ + 3t 2 ˆj ms −2 . If the particle is initially at
rest, find the velocity of the particle at time t = 2 s .
13. When a particle is projected vertically upwards
with an initial velocity of v0, it experiences an acceleration a = − g + kv 2 , where g is the acceleration
due to gravity, k is a constant and v is the velocity of the particle. Determine the maximum height
reached by the particle.
14. If the velocity v of a particle moving along a straight
line decreases linearly with its displacement from
20 ms −1 to a value approaching zero at x = 30 m ,
determine the acceleration of the particle when
x = 15 m and show that the particle never reaches
the 30 m displacement.
15. A particle is moving along a straight line with the
(
)
(
)
acceleration a = ( 12t − 3 t ) ms −2, where t is in
seconds. Determine the velocity and the position of
the particle as a function of time. Assume that, initially the particle starts from rest from x 0 = 15 m.
11/28/2019 7:04:00 PM
4.24 JEE Advanced Physics: Mechanics – I
GRAPHICAL INTERPRETATION OF SOME
QUANTITIES
⎛ Δx ⎞
As Δt → 0 , vav ⎜ =
→ vinst.
⎝ Δt ⎟⎠
Average Velocity
Geometrically, as
at P .
If a particle passes a point P ( xi ) at time t = ti and
reaches Q ( x f ) at a later time instant t = t f , its average velocity in the interval PQ is
vav =
xf
C
v<0
t
O
When the slope of the x -t graph is positive, v is positive (as at the point A in figure). At C , v is negative
because the tangent has negative slope. The instantaneous velocity at point B (turning point) is zero as
the slope is zero.
P
ti
B
A
Q
xi
t
tf
Instantaneous Acceleration
The derivative of velocity with respect to time is the
slope of the tangent in velocity time ( v -t ) graph.
Instantaneous Velocity
Consider the motion of the particle between the two
points P and Q on the x -t graph shown. As the
point Q is brought closer and closer to the point P ,
the time interval between PQ ( Δt , Δt ′ , Δt ′′ ,........ )
get progressively smaller. The average velocity for
each time interval is the slope of the appropriate dotted line ( PQ , PQ ′ , PQ ′′..... ) .
As the point Q approaches P , the time interval approaches zero, but at the same time the slope of
the dotted line approaches that of the tangent to the
curve at the point P .
x
Q″
Q′
P
ti Δt″
tf
v
a=0
a>0
a=0
O
t
a<0
a=0
t
Motion with Uniform Velocity
Consider a particle moving along x-axis with uniform
velocity u starting from the point x = xi at t = 0 .
Equations of x , v , a are
1. x ( t ) = xi + ut
Q
Tangent
O
v=0
Q(xf)
x
O
x
v>0
This expression suggests that the average velocity is
equal to the slope of the line (chord) joining the points
P and Q on the x -t graph.
PQ → tangent
Hence the instantaneous velocity at P is the slope of
the tangent at P in the x -t graph.
Δx x f − xi
= ( Slope of chord PQ )
=
Δt
t f − ti
P(xi)
Δt → 0 , chord
2. v ( t ) = u
3. a ( t ) = 0
So, x -t graph is a straight line of slope u through xi
As velocity is constant, v -t graph is a horizontal
line and a-t graph coincides with time axis because
a = 0 at all-time instants.
Δt′
Δt
04_Kinematics 1_Part 1.indd 24
11/28/2019 7:04:09 PM
Chapter 4: Kinematics I
x
x
e
p
Slo
xi
=u
pe
(i)
=u
y 2 = kx
y
O
t
u is positive
v
O
u is negative
v
t u
x
t
O
O
t
a
Positive velocity
u
(b) Parabolic Curve-Equation and Graphs
Slo
xi
4.25
(ii)
Negative velocity
y 2 = −kx
y
t
O
x
Conceptual Note(s)
(iii) x 2 = ky
Graph Identification
y
(a)Straight Line Equation, Graph, Slope (+ve, –ve,
zero slope)
If q is the angle at which a straight line is inclined to
the positive direction of x-axis and 0° ≤ θ < 180°,
θ ≠ 90°, then the slope of the line, denoted by m,
is defined by m = tanθ .
If q is 90° , m does not exist, but the line is parallel
to the y-axis.
If θ = 0 , then m = 0 and the line is parallel to the
x-axis.
For a straight line, the average slope is equal to
instantaneous slope, because a straight line has
constant slope.
Equation of Line: Slope-Intercept Form
y = mx + c is the equation of a straight line whose
slope is m and which makes an intercept c on the
y-axis i.e., the line cuts the y-axis at ( 0, + c ) .
m = slope = tanθ =
y
dy
dx
y
y
+ve slope
(0, c)
θ
c
04_Kinematics 1_Part 1.indd 25
x
x
(iv) x 2 = −ky
y
x
Where k is a positive constant.
(c) Equation of Parabola
CASE-1:
y = ax 2 + bx + c
For a > 0
The nature of the parabola will be, like that of the
nature of x 2 = ky
y
(0, c)
c
slope = 0
x
(0, c)
–ve slope
c
θ
x
x
11/28/2019 7:04:13 PM
4.26 JEE Advanced Physics: Mechanics – I
Interpretation of Some More Graphs
CASE-2:
a<0
Position vs Time Graph
The nature of the parabola will, like that of the
nature of x 2 = −ky .
y
(a) Zero Velocity: As position of particle is fixed at
all the times, so the body is at rest.
⇒
x
x
Graphs in Uniformly Accelerated
Motion (a ≠ 0)
(a) If acceleration a is constant, then a-t graph is a
horizontal line as shown.
a
a
Positive
acceleration
a
0
a =+
(b) For constant acceleration, v is a linear polynomial in terms of t . Hence v -t graph is a straight
line of slope a as shown.
v
Slo
pe
op
e
=
a
u
Sl
u
a is positive
0
v = u + at
t
(b) Uniform Velocity: Here tan θ is constant so,
dx
tan θ =
= constant
dt
⇒ Velocity of particle is constant.
x
Negative
acceleration
a=–
a
v
v = u – at
t
θ
t
(c) Non Uniform Velocity (Increasing with Time):
In this case, as time is i­ncreasing, θ is also
increasing.
⇒
=a
a is negative
0
0
dx
= tan θ also increases
dt
Hence, velocity of particle is increasing.
x
(c) For constant acceleration, x is a quadratic polynomial in terms of t . Hence x -t graph is a
parabola.
x = 1 at2
2
θ2
2
1
x
x
t
0
t
0
t
dx
= tan θ = tan 0° = 0
dt
Velocity of particle is zero
Slope =
0
v=0
θ1
θ2 > θ1
⇒ v2 > v1
t
(d) Non Uniform Velocity (Decreasing with Time):
In this case, as time increases, θ decreases.
0
x = ut + 1 at2
2
04_Kinematics 1_Part 1.indd 26
t
0 x = ut – 1 at2
2
t
⇒
dx
= tan θ also decreases
dt
11/28/2019 7:04:17 PM
4.27
Chapter 4: Kinematics I
Hence, velocity of particle is decreasing.
x
θ1
0
θ2 < θ1
⇒ v2 < v1
θ2
dv
= negative constant
dt
i.e., v -t graph is a straight line with negative
slope.
⇒
v
t
θ
Velocity vs Time Graph
(a) Zero Acceleration: Velocity is constant
⇒
tan θ = 0
⇒
dv
=0
dt
Acceleration vs Time Graph
(a) Constant Acceleration: tan θ = 0
⇒
Hence, acceleration is zero i.e., v -t graph is a
straight line parallel to time axis.
da
=0
dt
Hence acceleration is constant i.e., a-t graph is a
straight line parallel to time axis.
v
a
t
0
(b) Uniform Acceleration: tan θ is constant
⇒
t
0
dv
= constant
dt
i.e., v -t graph is a straight line with positive
slope.
v
t
0
(b) Uniformly
constant.
Increasing
Acceleration:
0° < θ < 90°
⇒
da
= tan θ = constant > 0
dt
⇒
q
is
tan θ > 0
Hence, acceleration is uniformly increasing with
time i.e., a-t graph is a straight line with positive
slope.
a
0
θ
t
(c) Uniform Retardation: Since θ > 90° , so tan θ is
constant and negative.
04_Kinematics 1_Part 1.indd 27
0
θ
t
11/28/2019 7:04:22 PM
4.28 JEE Advanced Physics: Mechanics – I
(c) Uniformly
θ > 90°
⇒
Decreasing
Acceleration:
Since
Hence, acceleration is constant throughout and is
equal to 8 ms −2 .
tan θ is constant and negative
a
da
⇒ dt = negative constant
Hence, acceleration is uniformly decreasing with
time i.e., a-t graph is a straight line with negative slope.
8
t
0
a
Conceptual Note(s)
θ
t
0
Illustration 24
The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw
velocity vs time and acceleration vs time graph.
x
x = 4t 2
t
0
Solution
x = 4t 2
⇒
v=
dx
= 8t
dt
Illustration 25
v
For a particle moving along x-axis, velocity-time
graph is as shown in figure. Find the distance travelled and displacement of the particle?
tan θ = 8
0
θ
t
Hence, velocity-time graph is a straight line having
slope i.e., tan θ = 8
a=
dv
=8
dt
04_Kinematics 1_Part 1.indd 28
(a) For uniformly accelerated motion ( a ≠ 0 ) , x-t
graph is a parabola (opening upwards if a > 0
and opening downwards if a < 0 ). The slope of
tangent at any point of the parabola gives the
velocity at that instant.
(b) For uniformly accelerated motion ( a ≠ 0 ) , v-t
graph is a straight line whose slope gives the
acceleration of the particle.
(c) In general, the slope of tangent in x-t graph is
velocity and the slope of tangent in v-t graph is
the acceleration.
(d) The area under a-t graph gives the change in
velocity.
(e) The area between the v-t graph gives the distance
travelled by the particle, if we take all areas as
positive.
(f) Area under v-t graph gives displacement, if areas
below the t-axis are taken negative.
v ms–1
8
8
0
2
4
6
10
t(s)
–5
11/28/2019 7:04:26 PM
Chapter 4: Kinematics I
4.29
x = −8t 2
Solution
Distance travelled = Area under v -t graph (taking all
areas as positive)
⎛ Distance ⎞ ⎛ Area of ⎞ ⎛ Area of ⎞
⎜⎝ travelled ⎟⎠ = ⎜⎝ trapezium ⎟⎠ + ⎜⎝ triangle ⎟⎠
Distance travelled =
⇒
Distance travelled = 32 + 10 = 42 m v=
⇒
a = −16 ms −2 Illustration 27
1
1
( 2 + 6 ) (8) + ( 4 ) (5)
2
2
⇒
dx
= −16t
dt
⇒
Draw displacement-time and acceleration-time graph
for the given velocity-time graph.
v
Displacement = Area under v -t graph (taking areas
below time axis as negative)
⎛ Area of ⎞ ⎛ Area of ⎞
Displacement = ⎜
−
⎝ trapezium ⎟⎠ ⎜⎝ triangle ⎟⎠
t1
1
1
( 2 + 6 ) (8) − ( 4)(5)
2
2
⇒
Displacement =
⇒
Displacement = 32 − 10 = 22 m
t2
t
t3
Solution
a
s
Hence, distance travelled is 42 m and displacement
is 22 m
Illustration 26
The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw
velocity vs time and acceleration vs time graph.
x
x = –8t2
Solution
Taking upward direction as positive so, downward
direction is taken as negative.
v
a
t
t
–16 ms–2
04_Kinematics 1_Part 1.indd 29
t2
t3
t
t2
t1
t3
t
Illustration 28
t
0
t1
The speed of a train increases at a constant rate a from
zero to v and then remains constant for an interval
and finally decreases to zero at a constant rate b. If
l is the total distance travelled by the train then find
the total time taken to complete the ­journey. At what
value of v is the time of travel shortest? What is the
value of the shortest time?
Solution
Let us first draw v -t graph showing the situation
discussed.
For accelerated Motion (O to A)
vmax = v = α t1
⇒
t1 =
v
…(1)
α
11/28/2019 7:04:30 PM
4.30 JEE Advanced Physics: Mechanics – I
Also, l1 =
v2
∵ v 2 − 0 2 = 2α l1 …(2)
2α
{
}
For Uniform Motion (A to B)
l2 = vt2
⇒
l
t2 = 2 …(3)
v
l
⇒
−
⇒
v=
v2
1⎛ 1 1⎞
+ ⎜ + ⎟ =0
2⎝ α β⎠
2lαβ
…(7)
α +β
So, tMIN =
v
l
2lαβ
α+β
+
1 2lαβ ⎛ 1 1 ⎞
+
2 α + β ⎜⎝ α β ⎟⎠
A
vmax = v
B
l2
l3
O
t1
t2
t3
C
t
For Decelerated Motion (B to C)
0 = v + ( − β ) t3
v2
∵ 0 2 − v 2 = 2 ( −β ) l3 …(5)
2β
{
}
If t be the total time of journey, then
t = t1 + t2 + t3
⇒
⇒
tMIN =
l (α + β )
l (α + β )
+
2αβ
2αβ
⇒
tMIN = 2
l (α + β )
=
2αβ
⇒
⎛ 1 1⎞
tMIN = 2l ⎜ + ⎟
⎝α β⎠
v
t3 = …(4)
β
Also, l3 =
⇒
l (α + β )
lαβ ⎛ α + β ⎞
+
2αβ
2 ( α + β ) ⎜⎝ αβ ⎟⎠
β
l1
⇒
tMIN =
⇒
α
⇒
{Substitute (7) in (6)}
t=
v
+
α
v
v
+
v
β
⇒
v l
v
v v
t= + −
−
+
α v 2α 2β β ⇒
t=
{∵ l1 + l2 + l3 = l }
α
O
{using (2) and (5)}
l v⎛ 1 1⎞
+
+
…(6)
v 2 ⎜⎝ α β ⎟⎠
For t to be MINIMUM, we must have
04_Kinematics 1_Part 1.indd 30
v
⎛ v2 v2 ⎞
v l ⎜
2β ⎟ v
⎟+ t = + − ⎜ 2α +
α v ⎝ v
v ⎠ β
dt
=0
dv
Please note that this time equals the time of journey
when the train just accelerated to attain a velocity v
and immediately decelerated to zero velocity after
covering a total distance l. Then
v
v l
t= + 2 +
α v β
l − ( l1 + l3 )
2l ( α + β )
αβ
β
t
t1
t2
tMIN = t1 + t2
⇒
tMIN =
⇒
⎛ 1 1⎞
tMIN = v ⎜ + ⎟ …(1)
⎝α β⎠
Also, l =
v v
+
α β
v2 v2
+
2α 2β
11/28/2019 7:04:39 PM
4.31
Chapter 4: Kinematics I
Since t1 + t2 + t3 + t4 + t5 =
2lαβ
⇒ v=
α +β
Substituting in (1), we get
⎛ 1 1⎞
tMIN = 2l ⎜ + ⎟
⎝α β⎠
⇒
100
⎛ 100 − 70 ⎞
⎛ 70 ⎞ 25
min
+ t2 + ⎜
⎟⎠ + t4 + ⎜⎝ ⎟⎠ =
⎝
a
a
a
3
⇒
t2 + t 4 +
200 25
=
min
a
3
⇒
t2 + t 4 +
200 25
=
hr …(1)
a
180
Illustration 29
Two road rally checkpoints A and B are located on
the same highway and are 12 km apart. The speed
limits for the first 8 km and the last 4 km of the section
of highway are 100 kmh −1 and 70 kmh −1 , respectively. Drivers must stop at each checkpoint and the
25
specified time between points A and B is
min.
3
Knowing that a driver accelerates and decelerates
at the same constant rate, determine the magnitude
of his acceleration if he travels at the speed limit as
much as possible.
Solution
Total time taken by the driver to go from A to B is
25
min . We must note here that a driver first acceler3
ates with acceleration a (say) for time t1 , moves with
uniform velocity of 100 kmh −1 for some time t2 (say)
and retards to a speed of 70 kmh −1 with deceleration
a (again) for a time t3 (say) to travel a total distance
of 8 km . Just after he completes 8 km , he maintains
his speed at 70 kmh −1 for a time t4 (say) and then
finally retards with deceleration a (again) for a time
t5 (say) to complete a further distance of 4 km . As
per this discussion, the v -t plot for the situation is
shown here.
70
D
M
E
O
04_Kinematics 1_Part 1.indd 31
1 ⎛ 100 ⎞
1 ⎛ 30 ⎞
⎜⎝
⎟⎠ ( 100 ) + 100t2 + ⎜⎝
⎟ ( 100 + 70 ) = 8
2 a
2 a ⎠
⇒
5000
2550
+ 100t2 +
=8
a
a
⇒
7550
+ 100t2 = 8 …(2)
a
Also, Area ( CDE ) = 4 km
⇒
⇒
t2
1 ⎛ 70 ⎞
70t4 + ⎜ ⎟ ( 70 ) = 4 km
2⎝ a ⎠
2450
+ 70t4 = 4 …(3)
a
From (2), (3), we get
t2 =
1 ⎛
7550 ⎞
1 ⎛
2450 ⎞
⎜⎝ 8 −
⎟⎠ and t4 =
⎜⎝ 4 −
⎟
100
70
a
a ⎠
Substituting in (1), we get
1 ⎛
7550 ⎞ 1 ⎛
2450 ⎞ 200 25
=
⎜⎝ 8 −
⎟⎠ + ⎜⎝ 4 −
⎟+
100
70
180
a
a ⎠
a
8
7550 4 2450 200 25
−
+
−
+
=
100 100 a 70 70 a
a
180 96 75.5 35 200 25
−
−
+
=
700
a
a
a
180
C
100
a
Area ( OABCM ) = 8 km
B
A
100
Further, area under the v -t graph gives displacement, so
⇒
v(kmhr –1)
25
min
3
30 t4
a
70
a
t(hr)
⇒
89.5
11
=
a
6300 ⇒
a = 51259 kmhr −2 ⇒
a = 3.96 ms −2 11/28/2019 7:04:51 PM
4.32 JEE Advanced Physics: Mechanics – I
Illustration 30
Solution
A particle starts from rest and traverses a distance
l with uniform acceleration, then moves uniformly
over a further distance 2l and finally comes to rest
after moving a further distance 3l under uniform
retardation. Assuming entire motion to be rectilinear
motion, find the ratio of average speed over the journey to the maximum speed on its way.
Since, s1 + s2 = 4 km and t1 + t2 = 4 min {given}
Also, t1 =
t2 =
vmax
and
α
vmax
β
Solution
v
Total Distance
vav =
Total Time
⇒
l + 2l + 3 l
vav =
t1 + t2 + t3
For OA, l =
⇒
t1 =
1
t1vmax
2
2l
A
vmax
t1
α
B
O
2l
l
O
M
3l
N
t2
t3
C
t
vmax
For AB, 2l = vm t2
⇒
t2 =
2l
vmax
For BC, 3l =
⇒
⇒
t3 =
1
t3 vmax
2
t1
s2
N
β
t2
B
t
⎛ 1 1⎞
t1 + t2 = vmax ⎜ + ⎟ …(1)
⎝α β⎠
and s2 =
1
vmax t1 …(2)
2
1
vmax t2 …(3)
2
Adding (2) and (3) we get,
6l
vmax = 2 km(min)−1
vmax
1 1
−2
+ = 2 km ( min )
α β
∴
v
3
Hence, av =
vmax 5
Illustration 31
A train starts from station A with uniform acceleration α for some distance and then goes with uniform
retardation b for some more distance to come to rest
at station B. The distance between station A and B is
4 km and the train takes 4 minute to complete this
1 1
journey, then find the value of
+ , where a and
α β
−2
β are in km ( min ) .
04_Kinematics 1_Part 1.indd 32
s1
Adding we get,
Now s1 =
6l
vav =
⎛ 10l ⎞
⎜⎝ v
⎟
max ⎠
A
vmax
v
INTERPRETATION OF GRAPHS OF VARIOUS
TYPES OF MOTION
Interpretation of x - t Graph
(a) The line on the graph representing x ( t ) is called
a World-Line.
(b) A point on the World Line is called an Event.
(c) The average slope of the World Line gives the
average velocity. Just join the two points with
a line and then find the angle which this line
(­joining the points) makes with the x-axis. If this
angle is θ , then
Average Velocity = vav = tan θ
11/28/2019 7:04:55 PM
Chapter 4: Kinematics I
(d) a-t graph can also be used to find the average
acceleration of the body, because
x
B
A
θ
t
vP = vinstantaneous = tan ϕ
x
v 2 − u2
2
(b) Slope of v 2 -x graph gives twice the acceleration
d ( 2)
v = 2a
i.e.
dx
(c) From v -x graph we can also find the instantaneous acceleration of a particle. Suppose we have
to find the acceleration aP at a point P on the
v -x curve. To calculate aP , we follow the steps
given below.
STEP-1: First find the velocity vP of the particle at
the point P .
Tangent at
point P
STEP-2: Then calculate the instantaneous slope
⎛ dv ⎞
⎜⎝
⎟ = tan θ of the curve at the same point.
dx ⎠ P
P
ϕ
vP = vInstantaneous = tan ϕ
t
Interpretation of v - t Graph
(a) The average slope of a curve in v ( t ) graph gives
average acceleration.
(b)The instantaneous slope of a curve at a point in
v ( t ) graph gives the instantaneous acceleration.
(c) The area under a curve in v ( t ) graph gives the
displacement of the body.
(d) v -t graph can also be used to find the average
velocity of the body, because
STEP-3: Finally, the acceleration at the point P is
the product of values calculated in Steps 1 and 2 i.e.
⎛ dv ⎞
aP = vP ⎜
.
⎝ dx ⎟⎠ P
Illustration 32
For an airplane to take-off it accelerates according to
the graph shown and takes 12 s to take-off from the
rest position. Calculate the distance travelled by the
airplane from t = 0 to t = 12 s .
Acceleration
ms–2
Δx Area under v -t graph
vav =
=
Δt
Time Interval
(a) The area under a curve in a-t graph gives the
change in the velocity of the body. So
Area under a-t graph = Δv = v f − v i
04_Kinematics 1_Part 1.indd 33
5
O
Interpretation of a - t Graph
Δv Area under a-t graph
=
Δt
Time Interval
(a) Area under a curve in a-x graph =
(d) The slope of an event gives the instantaneous
velocity. Just draw a tangent to the curve at
a point P and then find the angle which this
tangent makes with the x-axis. If this angle is
ϕ , then the instantaneous velocity at the point
P is
aav =
Interpretation to Other Graphs
Average velocity = tan θ
4.33
A
B
6
12
t(in s)
Solution
From O → A we have acceleration varying linearly
with time hence
11/28/2019 7:05:01 PM
4.34 JEE Advanced Physics: Mechanics – I
a−0 =
5
(t − 0 )
6
a=
⇒
dv 5
= t
dt 6 v1
∫
0
⇒
v1 =
t
dv =
v1 =
5
t dt
6
∫
0
5 2
t
12 Further v1 =
x1
1
0
x1 =
⇒
5
1.
2
1
2
x2 = v1 ( 6 ) + ( 5 )( 6 )
2
⇒
x2 = 180 m Now, let us plot v -t and s-t graphs of some standard
results. To draw the following graphs assume that the
particle has got either a one-dimensional motion with
uniform velocity or with constant acceleration.
6
5 3
t
36 0 Situation
v-t Graph
Uniform motion
s-t Graph
v
Uniformly accelerated
motion with u = 0 and
s = 0 at t = 0
t
s
s = 1 at 2
2
v = at
t
v
u
(i) Slope of s-t graph i.e. v is
constant.
(ii) In s-t graph s = 0 at t = 0.
s = vt
v
Uniformly accelerated
motion with u ≠ 0 but
s = 0 at t = 0
Interpretation
s
t
3.
⇒
GRAPHS
6
v = constant
2.
5
( 36 ) = 15 ms −1
12
x = 30 + 180 = 210 m
dx1
5
= t2
dt 12
0
S.No.
5 ( 3)
6 = 30 m
36
So, total distance travelled is
∫ dx = 12 ∫ t dt
⇒
x1 =
For A → B the acceleration is constant and have a
value of 5 ms −2 . Now velocity at point A is
5
t
6 ⇒
⇒
⇒
t
s
v = u + at
s = ut + 1 at 2
2
t
(i) u = 0, i.e., v = 0 at t = 0.
(ii) u = 0, i.e., slope of s-t graph
at t = 0, should be zero.
(iii) a or slope of v-t graph is
constant.
(i) u ≠ 0, i.e., v or slope of s-t
graph at t = 0 is not zero.
(ii) v or slope of s-t graph
gradually goes on
increasing.
t
(Continued)
04_Kinematics 1_Part 1.indd 34
11/28/2019 7:05:04 PM
Chapter 4: Kinematics I
S.No.
Situation
4.
Uniformly accelerated
motion with u ≠ 0 and
s = s0 at t = 0
v-t Graph
s-t Graph
v
s0
t
5.
Uniformly retarded
motion till velocity
becomes zero
Interpretation
(i) s = s0 at t = 0
s
v = u + at
u
v
s = s0 + ut + 1 at 2
2
t
(i) Slope of s-t graph at t = 0
gives u.
(ii) Slope of s-t graph at t = t0
becomes zero.
(iii) In this case u can’t be zero.
s
u
v = u – at
t0
6.
Uniformly retarded then
accelerated in opposite
direction.
t
v
t0
t
s
u
t0
t
0
Conceptual Note(s)
v
s
t
t
(b) At one time, two values of velocity or displacement are not possible. Hence, the following
graphs are not acceptable:
04_Kinematics 1_Part 1.indd 35
t
t0
(i) At time t = t0, v = 0 or slope
of s-t graph is zero.
(ii) In s-t graph slope or
velocity first decreases then
increases with opposite
sign.
v
(a) Slopes of v-t or s-t graphs can never be infinite
at any point, because infinite slope of v-t graph
means infinite acceleration. Similarly, infinite
slope of s-t graph means infinite velocity. Hence,
the following graphs are not possible:
4.35
s
v1
s1
v2
s2
t0
t
t0
t
(c) Different values of displacements in s-t graph corresponding to given v-t graph can be calculated
just by calculating areas under v-t graph. There is
no need of using equations like v = u + at, etc.
Illustration 33
A particle starting from rest undergoes an acceleration that increases linearly with time. Estimate the
particle’s velocity n second after the start and the
distance moved by the particle in these n second.
Draw the graph showing the variation of acceleration, velocity and distance with time.
11/28/2019 7:05:07 PM
4.36 JEE Advanced Physics: Mechanics – I
Solution
a
a = kt
⇒
⇒
a = kt
dv
= kt
dt
dv = ktdt v
⇒
∫ dv = k∫ t dt
0
2
dx kt
=
dt
2 ⇒
dx =
kt 2
2
k
v
t
O
x
x=
2
kt 3
6
∫
3
⇒
t
0
a= 0
dv v
=
dt 5
⇒
∫ dx = 2 ∫ t dt
0
⇒
v=
kt 2
dt
2
x
⇒
v
2
⇒
3 x
+
5 25 The corresponding a-x graph is shown here
From equation (1), we have
2
a=
i.e., the a-x graph is a straight line, where at x = 0 ,
3
a = ms −2 = 0.6 ms −2 and at s = 60 m , a = 3 ms −2
5
For s > 60 m , we have v = constant
⇒
kt
kn
=
2
2 dx
Further v =
dt
v=
t
O
t
0
⇒
⇒
⇒
a(ms–2)
t
dv 1
dt
=
v 5
∫
0
⎛ v⎞ t
ln ⎜ ⎟ =
⎝ 3⎠ 5
⇒
kt 3 kn3
x=
=
3
6 t
⇒
0.6
B
120
x(m)
∵ v=
dx
dt
60
{
t1
∫ dx = 3∫ e dt t5
0
O
3
v = 3et 5 60
0
}
60 = 15 ( e t1 5 − 1 ) Illustration 34
⇒
The velocity-displacement (v-x) graph describing the
motion of a motorcycle is shown in figure. Construct the
a-x graph of the motion and determine the time needed
for the motorcycle to reach the position x = 120 m.
Time taken to travel next 60 m with speed 15 ms −1
will be t2 given by
v(ms–1)
60
120
x(m)
Solution
For 0 < x ≤ 60 m , we have
12
x
x+3 = 3+
60
5
dv ⎛ 1 ⎞ dx 1
1⎛
x⎞ 3 x
=⎜ ⎟
= (v) = ⎜ 3 + ⎟ = +
…(1)
dt ⎝ 5 ⎠ dt 5
5⎝
5 ⎠ 5 25
04_Kinematics 1_Part 1.indd 36
60
=4s
15
Total time = t1 + t2 = 8 + 4 = 12 s
Illustration 35
3
v=
t2 =
∴
15
t1 = 8 s The acceleration-displacement ( a − x ) graph of a
particle moving in a straight line is as shown. Assume
the particle to start from rest, find the velocity of the
particle when displacement of the particle is, 12 m .
a(ms–2)
4
2
2
8
10
12
x(m)
11/28/2019 7:05:14 PM
Chapter 4: Kinematics I
sOLuTION
Since a =
dv
dt
⇒
dv
v
=a
dx
⇒
vdv = adx
⇒
v2
= area under a-x graph
2
⇒
v = 2 ( area under a-x graph )
The area A under the a-x graph is given by
A=
v
⇒
4.37
∫ vdv =∫ adx
0
1
1
1
( 2 )( 2 ) + (6)(2) + ( 2 + 4 )( 2 ) + (2)( 4)
2
2
2
⇒
A = 2 + 12 + 6 + 4 = 24 m 2 s −2
⇒
v = 2 × 24 = 4 3 ms −1
Test Your Concepts-IV
Based on Graph
1. A two stage missile is fired vertically from rest with
the acceleration shown. In 15 s the first stage burns
out and the second stage ignites. Plot the v-t and
s-t graphs which describe the two-stage motion of
the missile for 0 ≤ t ≤ 20 s.
(Solutions on page H.77)
3. The car is moving along a straight road with the
speed described by the v-x graph. Construct the
a-x graph.
v(ms–1)
a(ms–2)
75
25
18
v=
v=5 x
–0
.2x
+1
20
15
15
20
t(s)
2. The particle moves rectilinearly with the velocity
described by the graph. Construct the acceleration-displacement graph.
v(ms–1)
O
04_Kinematics 1_Part 1.indd 37
525
x(m)
4. A motorcyclist starting from rest travels along a
straight road and for 0 < t ≤ 10 s, it has an acceleration as shown. Draw the v-t graph that describes
the motion and find the distance travelled in 10 s.
a(ms–2)
13
10
4
225
6
v=x+7
v = 2x + 4
3
6
x(m)
1
a = t2
6
6
10
t(s)
11/28/2019 7:05:18 PM
4.38 JEE Advanced Physics: Mechanics – I
5. The dragster starts from rest and travels along a
straight track with an acceleration-deceleration
described by the graph. Construct the v-x graph for
0 ≤ x ≤ x 0 and determine the distance x0 travelled
before the dragster again comes to rest.
a(ms–2)
5
a=
v(ms–1)
50
40
25
x+
9. The v-x graph for an airplane travelling on a straight
run-way is shown. Determine the acceleration of
the plane at x = 50 m and x = 150 m. Draw the
a-x graph.
5
1
0.
x0
200
x(m)
–15
6. A particle travels along a curve defined by the equation x = ( t 3 − 3t 2 + 2t ) m, where t is in s­econds.
Draw the x-t, v-t and a-t graphs for the particle for
0 ≤ t ≤ 3 s.
7. A truck is travelling along the straight line with a
velocity described by the graph. Construct the a-x
graph for 0 ≤ x ≤ 1500 m.
O
100
200
10. Starting from rest from x = 0, a boat travels in a
straight line with an acceleration as shown by the
a-x graph. Determine the boat’s speed at x = 40 m,
90 m and 200 m.
a(ms–2)
4
2
v(ms–1)
50
O
150
250
x(m)
11. The acceleration of particle moving rectilinearly,
varies with time as shown.
75
a(ms–2)
v = 0.6 s3/4
625
1500
x(m)
8. The jet plane starts from rest from x = 0 and is subjected to the acceleration shown. Determine the
speed of the plane when it has travelled 60 m.
a(ms–2)
1
t(s)
–2
22.5
(a) Find an expression for velocity in terms of t.
(b)Calculate the displacement of the particle in
the interval from t = 2 s to t = 4 s . Assume
that v = 0 at t = 0.
150
04_Kinematics 1_Part 1.indd 38
x(m)
x(m)
11/28/2019 7:05:20 PM
Chapter 4: Kinematics I
12. Acceleration-time graph of a particle moving in a
straight line is shown in figure. Velocity of particle
at time t = 0 is 2 ms −1. Find velocity at the end of
fourth second.
13. Velocity-time graph of a particle moving in a
straight line is shown in ­
figure. Plot the corresponding displacement-time graph of the particle
if at time t = 0, displacement s = 0.
v(ms–2)
10
Step-2
The quantities ( i.e. u , v , a and s ) along the initial
direction of motion will be taken as positive whereas
opposite to the initial direction of motion are taken as
negative.
For example
(a) For the particle dropped from the tower of height h,
we have taken downward direction (the direction of
initial motion) as positive. So, in this case we have
u = 0, v = + v, a = + g and s = +h
as all are in the downward direction.
(b) For the particle thrown vertically upwards from a
tower of height h, we have taken upward direction
(the direction of initial motion) as positive. So, in this
case, since
• initial velocity u is upwards, so u = +u,
• acceleration due to gravity is downwards, so
a = − g,
04_Kinematics 1_Part 1.indd 39
A
B
2
4
O
D
8
6
t(s)
• particle eventually reaches the ground, so the displacement of the particle is downwards and hence
s = Δy = −h.
Step-3
Now apply the equations of rectilinear motion of the
particle i.e.
Step-1
Take initial direction of motion of the particle as
positive.
For example
(a) If a particle is dropped from a tower, then we can
take downward direction (the direction of initial
motion) as positive.
(b) If a particle is thrown vertically upwards from a tower,
then we can take upward direction (the direction of
initial motion) as positive.
C
20
VERTICAL MOTION UNDER GRAVITY
Vertical motion under gravity, is the case of motion
of a particle moving rectilinearly under the influence
of gravity. To solve problems involving motion of a
particle under gravity, we make use of the following
steps for the sake of convenience and accuracy.
4.39
v = u + at ,
s = ut +
v 2 − u2 = 2 as ,
s=
snth = u +
1 2
at
2
u+v
t
2
1
a ( 2n − 1 )
2
and substituting the values as calculated in Step 2, to
get the desired solution to the problems as discussed
below.
For example
(a) For a body dropped ( u = 0 ) from a height h, the
equation of motion are
1
v = gt, h = gt 2 , v = 2 gh
2
Initial velocity in case of dropping is zero.
(b) For a body thrown downward with initial velocity u
from a height h, the ­equations of motion are
1
h = ut + gt 2 ,
2
v = u + gt,
g
u
v = u2 + 2 gh
h
v
Take downwards as positive
11/28/2019 7:05:23 PM
4.40 JEE Advanced Physics: Mechanics – I
(c) For a body launched up from the ground, with initial
velocity u, the equations of motion before the particle attains maximum height are
1
h = ut − gt 2 ,
2
v = u − gt,
v = u2 − 2 gh
Illustration 36
A ball is thrown upwards from the ground with an
initial speed of u . At two instants of time, having an
interval of 6 s , the ball is at a height of 80 m from the
ground. Find u . Take g = 10 ms −2 .
Solution
METHOD I
Consider the upward direction as positive, then
g
v
u = + u ms −1 , a = g = −10 ms −2 and s = +80 m
Since, s = ut +
h
1 2
at , so we get
2
s = 80 m
80 = ut − 5t 2
Take upwards as positive
When the particle returns to the ground, then s = 0
and hence we have
1
0 = ut − gt 2
2
⇒ t = Time of Flight =
2u
g
(d) For a body launched up from a tower of height h,
taking upward direction as positive, then we have
s = −h,
u = +u,
⇒
5t 2 − ut + 80 = 0 ⇒
t=
and
u − u2 − 1600
10
⇒
u2 − 1600
=6
5
⇒
u2 − 1600 = 30 ⇒
⇒
⇒
⇒
v
So, the equations of motion are written as
and
−h = ut +
1
( −g )t2
2
u2 = 2500 u = ±50 ms −1 u = 50 ms −1 B
Since tA→ A′ = 6 s ⇒
tA→B = tB→ A′ = 3 s For A → B
3s
3s
v
A
A′
0 = v + ( −10 )( 3 )
⇒
04_Kinematics 1_Part 1.indd 40
u2 − 1600 = 900 METHOD II
v = u + ( −g )t
v 2 − u2 = 2 ( − g ) ( −h )
u
u + u2 − 1600 u − u2 − 1600
−
=6
10
10
g
h
–ve
u + u2 − 1600
10
Now it is given that these two instants, when the ball
is at 80 m from the ground have an interval of 6 s in
between. So,
a = −g
u
+ve
v = 30 ms −1 For O → A
v 2 − u2 = 2 ( −10 )( 80 )
v
80 m
u
O
O′
11/28/2019 7:05:29 PM
4.41
Chapter 4: Kinematics I
⇒
900 − u2 = −1600 ⇒
u2 = 900 + 1600 ⇒
u = 2500 = 50 ms −1 Illustration 38
A particle is dropped from a tower is found to travel
45 m in the last second of its journey. Calculate the
height of the tower.
Solution
Illustration 37
A particle is dropped from height 100 m and another
particle is projected vertically up with velocity
50 ms −1 from the ground along the same line. Find
out the position where two particle will meet?
Solution
Let the upward direction be taken as positive.
Let the particles meet at a distance y from the ground.
Let the total time of journey be n seconds. Then
a
snth = u + ( 2n − 1 )
2
10
⇒ 45 = 0 + ( 2n − 1 )
2
⇒ n= 5 s
So, 5th second is the last second of motion. Hence
height of tower is given by
y0 = +100 m
y = 100 m
u=0
A
u = 0 ms–1
Since y = y0 + ut +
⇒
⇒
a = −10 ms −2
1 2
at
2
Illustration 39
y=0m
1
× 10 × t 2
2
u = 50 ms–1
B
⇒
y = 100 + 0 ( t ) −
⇒
y = 100 − 5t 2 …(1)
For particle B
A particle is projected vertically upwards. Prove that
3
it will be at
of its greatest height at time which are
4
in the ratio 1 : 3 .
Solution
For the particle projected upwards with an initial
velocity u , the greatest height attained is
y0 = 0 m
u = +50 ms −1
a = −10 ms −2
u2
2g
Let t be the time when the particle is at a height
H=
1
× 10 × t 2
2
⇒
y = 50 ( t ) −
⇒
y = 50t − 5t 2 …(2)
According to the problem
h=
t= 2 s
Putting t = 2 s in equation (1), we get
y = 100 − 20 = 80 m
Hence, the particles will meet at a height 80 m above
the ground.
04_Kinematics 1_Part 1.indd 41
3 H 3 ⎛ u2 ⎞
= ⎜
4
4 ⎝ 2 g ⎟⎠
Using the formula s = ut +
1 2
gt , we get
2
3 ⎛ u2 ⎞
1
= ut + gt 2
4 ⎜⎝ 2 g ⎟⎠
2
50t − 5t 2 = 100 − 5t 2
⇒
1 2
gt
2
1
2
h = ( 10 )( 5 )
2
h = 125 m h=
For particle A
⇒
t2 −
2u
6u 2
t+ 2 = 0
g
8g
Solving for t , we have
t=
⎛ 4u 2 3u 2 ⎞
2u
± ⎜ 2 − 2 ⎟
g
g ⎠
⎝ g
2
=
u u
±
g 2g
11/28/2019 7:05:35 PM
4.42 JEE Advanced Physics: Mechanics – I
Taking negative sign, t1 =
u
2g
Taking only positive sign, t2 =
⇒
⎛ u ⎞
t1 ⎜⎝ 2 g ⎟⎠
=
= 1: 3
t2 ⎛ 3 u ⎞
⎜⎝ 2 g ⎟⎠
Illustration 41
A stone is dropped from a balloon going up with a uniform velocity of 5 ms −1 . If the balloon was 60 m high
when the stone was dropped, find the height of balloon when the stone hits the ground. Take g = 10 ms −2.
3u
2g
Solution
s = ut +
Illustration 40
A body projected vertically upwards from the top
of a tower reaches the ground in t1 second. If it is
projected vertically downwards from the same point
with same speed, it reaches the ground in t2 second.
If it is just dropped from the top, it reaches the ground
in t second. Prove that t = t1t2 .
Let h be the height of the tower. Taking downward
direction as positive, we get
1
for the body projected upward, h = −ut1 + gt12…(1)
2
1
for the body projected downward, h = ut2 + gt22…(2)
2
for the freely falling body, h = 0 +
1 2 1 2
gt = gt …(3)
2
2
Multiplying equation (1) by t2 , we get
1 2
gt1 t2
2
Multiplying equation (2) by t1 , we get
ht2 = −ut1t2 +
⇒
⇒
1
gt1t2 …(4)
2
1 2 1
1
gt = gt1t2 ∵ h = gt 2 from equation (3)
2
2
2
h=
⇒
t 2 = t1t2 ⇒
t = t1t2
04_Kinematics 1_Part 1.indd 42
⇒
−60 = 5t − 5t 2 +ve
2
⇒
5t − 5t − 60 = 0 ⇒
t 2 − t − 12 = 0 ⇒
t 2 − 4t + 3t − 12 = 0 60 m –ve
( t − 4 )( t + 3 ) = 0 t = 4 s
Height of balloon from ground at this instant is
h = 60 + ( 4 ) ( 5 )
⇒
h = 80 m Conceptual Note(s)
As the particle is detached from the balloon it is having the same velocity as that of balloon, but it is moving under the influence of gravity, so a = -g (Because,
g is acting downwards and we have taken upward
direction as positive).
A particle is projected vertically upwards from a
point A on the ground. It takes t1 second to reach
a point B at a height h from A but still continues
to move up. If it takes further t2 second from B to
ground again, then show that
1 2
1
1
gt1 t2 + gt22 t1 = gt1t2 ( t1 + t2 )
2
2
2
{
1
−60 = 5 ( t ) + ( −10 ) t 2
2
Illustration 42
1 2
gt2 t1
2
Adding (1) and (2), we get
ht1 = ut1t2 +
h ( t1 + t2 ) =
⇒
⇒
⇒
Solution
1 2
at
2
}
1
gt1t2
2
2
g ( t1 + t2 )
(b) maximum height reached is
and
8
h
(c) the velocity of the particle at a height
is
2
g 2 2 12
.
t1 + t2
2
(a) h =
(
)
11/28/2019 7:05:41 PM
4.43
Chapter 4: Kinematics I
Solution
Illustration 43
(a) Let the particle be projected upwards with a
velocity u . Suppose t1 and t2 be the times taken
for the motion from A to B and for the motion
from B to C and then from C to A respectively.
A ball is dropped from a height of 80 m on a floor. At
each collision with the floor, the ball loses one-tenth
of its speed. Plot the speed-time graph of its motion
⎛ 2u ⎞
∴ Time of flight = ( t1 + t2 ) = ⎜
⎝ g ⎟⎠
g
( t1 + t2 )
2
For the motion AB ,
1
h = ut1 − gt12
2
g
1 2 1
⇒ h = 2 ( t1 + t2 ) t1 − 2 gt1 = 2 gt1t2
⇒
u=
between t = 0 to 11.2 s
Solution
Just Before First Collision
Time taken by the ball to fall through a height of 80 m
is obtained as follows:
u=0
h=
1 2
gt
2
80 =
1
× 10 × t 2
2
t=
2 × 80
=4s
10
C
⇒
B(t = t1)
( Take g = 10 ms−2 ) .
Now, v ( t ) = gt
h
⇒
From time t = 0 to t = 4 s , v ( t ) = gt = 10t
A(t = 0)
⇒
(b) Maximum height reached
⇒
AC = H =
AC =
1 ⎡g
u2
⎤
=
( t1 + t2 ) ⎥
2 g 2 g ⎢⎣ 2
⎦
g
( t1 + t2 )2
8
(c) Let v be the velocity at a height
⇒
v2 =
g
1
( t1 + t2 )2 − g × gt1t2
4
2
⇒
v2 =
g2 ⎡
( t1 + t2 )2 − 2t1t2 ⎤⎦
4 ⎣
⇒
⇒
2
g
v =
t12 + t22
4
g 2 2
v=
t1 + t2
2
04_Kinematics 1_Part 1.indd 43
2
(
)
2
h
, then
2
⎛ h⎞
v 2 = u2 − 2 g ⎜ ⎟ = u2 − gh
⎝ 2⎠
2
v ( 4 ) = 10 × 4 = 40 ms −1 v(t ) ∝ t In this duration speed increases linearly with time t
from 0 to 40 ms −1 during the downward motion of
the ball and this speed-time variation has been shown
by straight line OA in figure.
Speed (ms–1)
40
36
O
A
B
A: Just before first collision
B: Just after first collision
C: At maximum height
D: Just before second collision
D
C
4
7.6
11.2
O → A: Downward motion
B → C: Upward motion
D → C: Downward motion
t(s)
11/28/2019 7:05:47 PM
4.44 JEE Advanced Physics: Mechanics – I
Just After First Collision
At first collision with the floor
1
Speed lost by ball =
× 40 = 4 ms −1
10
Thus, the ball rebounds with a speed of
40 − 4 = 36 ms −1 . For the further upward motion, the
speed at any instant t is given by
v ( t ) = v ( 0 ) − gt = 36 − 10 × t
Since v = u + gt
⇒ 0 = 100 − 10 × t
So, time taken to reach highest point, is
100
t=
= 10 s
10
The ball will return to the ground at T = 2t = 20 s.
(b) Corresponding velocity-time graph of the ball is
shown in figure
v(ms–1)
Now, the speed decreases linearly with time and
becomes zero after time
+100
36
= 3.6 s
10
Thus, the ball reaches the highest point again after
time t = 4 + 3.6 = 7.6 s from the start. Straight line
BC represents the speed-time graph for this upward
motion.
A
t=
Just Before Second Collision
At highest point, speed of ball is zero. It again starts
falling. At any instant t its speed is given by
O
–50
B
0
5
C
10
15
t(s)
20
D
–100
(i) Maximum height ( H ) attained by the ball is
equal to the Area of ΔAOB
Again, the speed of the ball increases linearly with
time t from 0 to 36 ms −1 (initial speed of the previous
upward motion) in the next time-interval of 3.6 s. Total
time taken from the start is t = 4 + 3.6 + 3.6 = 11.2 s .
This part of motion has been shown by straight line
CD .
1
× 10 × 100 = 500 m
2
(ii) Height attained after 15 s is
h = Area of DAOB + Area of DBCD
1
⇒ h = 500 + 2 ( 15 − 10 ) × ( −50 )
⇒ h = 500 − 125 = 375 m
Illustration 44
Illustration 45
v ( t ) = 10t
A ball is thrown upward with an initial velocity of
100 ms −1 .
(a) Calculate the time taken by the ball to return to
the point of launch.
(b) Draw velocity-time graph for the ball and find
from the graph
(i) the maximum height attained by the ball and
(ii) height of the ball after 15 s.
−2
Take g = 10 ms .
Solution
(a) Taking upward direction as positive, we get
u = 100 ms −1 , g = −10 ms −2
At highest point i.e., the point of reversal of
velocity, v = 0
04_Kinematics 1_Part 1.indd 44
⇒
H=
A stone is dropped from the top of a cliff of height h . n
second later, a second stone is projected downwards
from the same cliff with a vertically downward velocity u . Show that the two stones will reach the bottom
2
2
of the cliff together, if 8 h ( u − gn ) = gn2 ( 2u − gn ) .
What can you say about the limiting value of n .
Solution
The time taken by the first stone to reach the bottom
2h
g
According to the given problem, the second stone is
projected n second later. The two stones will reach
the bottom together, if the second covers the same
2h
distance in time t =
−n
g
of the cliff is
11/28/2019 7:05:51 PM
Chapter 4: Kinematics I
Since, h = ut +
If t be the time taken by the body to reach the height
980 m , then from the equation v = u + gt , we have
1 2
gt
2
⎞
⎛ 2h
⎞ 1 ⎛ 2h
Hence, h = u ⎜
− n⎟ + g ⎜
− n⎟
⎠
⎝ g
⎠ 2 ⎝ g
⇒
⇒
⇒
h = ( u − gn )
n g⎞
2h ⎛
+ ⎜ h − un +
⎟
⎝
g
2 ⎠
2 h = ( u − gn )
2h n2 g
+
− un
g
2
⇒
gn2 ⎛ ( 2u − gn ) ⎞
h=
8 ⎜⎝ ( u − gn ) ⎟⎠
8 h ( u − gn ) = gn2 ( 2u − gn )
2h
−n≥ 0
g
⇒
2h
g
n≤
2h
g
Illustration 46
Two bodies are thrown vertically upward, with the
same initial velocity of 98 ms −1 but 4 s apart. How
long after the first one is thrown will they meet?
Let ymax be the maximum height at which the velocity of first body reduces to zero.
Using the equation v 2 = u2 + 2 gs , we have
2
0 = ( 98 ) − 2 × ( 9.8 ) × ymax
04_Kinematics 1_Part 1.indd 45
∵ s = ut +
1 2
gt
2
}
1
× 9.8 × t 2
2
and ( 78.4 − y ) = ( 39.2 ) t −
1
× 9.8 × t 2
2
Solving the two equations, we get
t=2s
Hence the two bodies shall meet 10 + 2 = 12 s after
the first body is thrown.
Illustration 47
Solution
ymax =
{
Now the two bodies will meet during the downward
journey of the first and the upward journey of the
­second. Let the two bodies meet after a time t measured from above moment. The first body is coming
down (initial velocity zero) and let it covers a distance
y in t second. The second body is moving upward
and covers a distance ( 78.4 − y ) in t second. Using
1
y = ut + gt 2 for two bodies, we have
2
y=
So the limiting value of n is nMAX =
⇒
1
2
y = ( 98 × 6 ) − ( 9.8 ) × ( 6 ) 2
y = 588 − 176.4 = 411.6 m Δy = 490 − 411.6 = 78.4 m
2
2h
− n , so we must have t ≥ 0
g
⇒
The height y reached by the body is
At this moment, the distance between the first and
second body is
Hence, the two stones will reach the bottom together,
if the above condition is satisfied. Also, since
t=
v = 98 − ( 9.8 ) × 6 = 39.2 ms −1
⇒
2
2
0 = 98 − ( 9.8 ) t
t = 10 s Now the second body which was thrown after 4 sec
has been moving upward for 6 sec. The velocity v
acquired by this body is given by
2
Solving it, we get
⇒
2
⎞
2h
1 ⎛ 2h
2h
− un + g ⎜
− 2n
+ n2 ⎟
g
2 ⎝ g
g
⎠
h=u
4.45
( 98 )2
= 490 m
2 × ( 9.8 )
A particle is dropped from the top of a tower h m
high and at the same moment another particle is projected upwards from the bottom. They meet when the
h
upper one has descended a distance . Show that
n
the velocities of the two when they meet are in the
ratio 2 : ( n − 2 ) and that the initial velocity of the parngh
ticle projected up is
.
2
11/28/2019 7:05:57 PM
4.46 JEE Advanced Physics: Mechanics – I
Solution
The situation is shown in figure.
B
u1 = 0
h/n
v1
A
1⎞ ⎤
⎡n
⎛
v22 = gh ⎢ − 2 ⎜ 1 − ⎟ ⎥
⎝
n⎠ ⎦
⎣2
⇒
⎛ n 2 − 4n + 4 ⎞
2⎞
⎛n
v22 = gh ⎜ − 2 + ⎟ = gh ⎜
⎟⎠
⎝2
⎝
n⎠
2n
⇒
h
(h – h/n)
⇒
v2
u
Let the two particles meet at A after a time t
For first particle,
For second particle,
h⎞
1
⎛
⎜ h − ⎟ = ut − gt 2 …(2)
⎝
⎠
n
2
Adding equations (1) and (2), we get
h = ut
⇒
t=
h
…(3)
u
Substituting the value of t from equation (3) in equation (1), we get
h 1 ⎛ h⎞
= g⎜ ⎟
n 2 ⎝ u⎠
⇒
⇒
2
u2 =
1
ngh
2
u=
ngh
…(4)
2
Velocity v1 of first particle at A is given by
⎛ h⎞
⎛ h⎞
v12 = u12 + 2 g ⎜ ⎟ = 0 + 2 g ⎜ ⎟ …(5)
⎝ n⎠
⎝ n⎠
Velocity v2 of second particle at A is given by
h⎞
⎛
v22 = u2 − 2 g ⎜ h − ⎟
⎝
n⎠
⇒
v22 =
1
1⎞
⎛
ngh − 2 gh ⎜ 1 − ⎟
⎝
2
n⎠ 04_Kinematics 1_Part 1.indd 46
…(6)
2 gh
4 gh
n
=
=
2
2
2
(n − 2)
v2
gh ( n − 2 )
gh
2n
v1
2
=
v2 n − 2
v12
Ground
h 1 2
= gt …(1)
n 2
2n
Dividing equation (5) by equation (6), we get
G
v22 = gh
( n − 2 )2
⇒
Illustration 48
A parachutist after bailing out falls 50 m , without
friction. When the parachute opens, he decelerates
downwards with 2 ms −2 . He reaches the ground
with a speed of 3 ms −1 .
(a) How long is the parachutist in the air?
(b) At what height did he bail out?
Solution
For fall without friction, i.e., free fall h = 50 m , u = 0 ,
g = 9.8 ms −2 and t = t1
1
Using h = ut + gt 2 , we have
2
1
50 = 0 + × 9.8 × t12
2
⇒
t1 =
50 × 2
= 3.16 s
9.8
The velocity after falling through 50 m may be
obtained by using the formula
v 2 = u2 + 2 gh
⇒
v 2 = 0 + 2 × 9.8 × 50 ⇒
v = ( 2 × 9.8 × 50 ) ⇒
v = 31.3 ms −1 Taking downward direction as positive, the initial
velocity is 31.3 ms −1 and final velocity is 3 ms −1 .
11/28/2019 7:06:03 PM
Chapter 4: Kinematics I
Acceleration a = −2 ms −2
Let time be t2 . Now using v = u + at, we have
3 = 31.3 − 2t2
⇒
t2 =
4.47
If distance travelled be h , then h = ut where
31.3 + 3 34.3
u=
=
= 17.15 ms −1 and t = 14.15 s
2
2
⇒ h = 17.15 × 14.15 = 242.7 m
Now, total time = t = 3.19 + 14.15 = 17.34 s
28.3
= 14.15 s
2
and total height = h = 50 + 242.7 = 292.7 m
Test Your Concepts-V
Based on Motion under Gravity
1. Can you think of examples where velocity of a
particle is
(a) in opposite direction to the acceleration
(b) zero but its acceleration is not zero
(c) perpendicular to the acceleration
2. As a body is projected to a high altitude above
the earth’s surface, the variation of the acceleration of gravity with respect to altitude y must be
taken into account. Neglecting air resistance,
this acceleration is determined from the formula
⎡ R2 ⎤
a = − g0 ⎢
, where g0 is the constant grav2 ⎥
⎢⎣ ( R + y ) ⎥⎦
itational acceleration at sea level, R is the radius
of the earth and the positive direction is measured upward. If g0 = 9.81 ms −2 and R = 6356 km ,
determine the minimum initial velocity at which a
projectile should be shot vertically from the earth’s
surface so that it does not return to the earth.
3. A particle is projected vertically upwards and t
second after another particle is projected upwards
with the same initial velocity. Prove that the part u
ticles will meet after a lapse of ⎛⎜ + ⎞⎟ second
⎝ 2 g⎠
from the instant of projection of the first particle.
What are the velocities of the particles when they
meet?
4. A ball is projected vertically upwards with a velocity of 100 ms −1. Find the speed of the ball at half
the maximum height. Take g = 10 ms −2 .
5. A particle is projected vertically upwards from the
ground at time t = 0 and reaches a height h at
t = T . Show that the greatest height of the particle
2
gT 2 + 2h )
(
is
.
8 gT 2
04_Kinematics 1_Part 1.indd 47
(Solutions on page H.81)
6. A rocket is fired vertically upwards with a net
acceleration of 4 ms −2 and initial velocity zero.
At t = 5 s its fuel is finished and it decelerates with
g. At the highest point its velocity becomes zero.
Then it accelerates downwards with acceleration g
and returns to ground. Plot velocity-time and displacement-time graphs for the complete journey.
Take g = 10 ms −2 .
7. A stone is dropped from the top of a tall cliff and
n second later another stone is thrown vertically
downwards with a velocity u ms −1 . How far below
the top of the cliff will the second stone overtake
the first?
8. A body of mass m is thrown straight up with a
velocity u0. Find the velocity u′ with which the
body comes down if the air drag equals cu2 where
c is a constant and u is the velocity of the body.
9. To test the quality of a tennis ball it is dropped onto
the floor from a height of 4 m and it rebounds to a
height of 2 m. If the ball is in contact with the floor
for a duration of 12 ms. Calculate the average acceleration during that contact? Take g = 9.8 ms −2 .
10. A rocket is fired vertically from rest and ascends with
constant net vertical acceleration of 300 msec −2
for 1 minute. Its fuel is then all used up and it continues as a free particle in the gravitational field of
the earth. Find
(a) maximum height reached
(b) the total time elapsed from take-off until the
rocket strikes the earth.
11. A football is kicked vertically upward from the
ground and a student gazing out of the window
sees it moving upwards past her at 5 ms −1. The
window is 15 m above the ground. Air resistance
may be ignored.
11/28/2019 7:06:07 PM
4.48 JEE Advanced Physics: Mechanics – I
(a) How high does the football go above ground?
(b)How much time does it take to go from the
ground to its highest point?
Take g = 10 ms −2 ?
12. Two balls of same mass are shot upward one
after another at an interval of 2 second along the
same vertical line with the same initial velocity of
39.2 msec −1. Find the height at which they collide.
Take g = 9.8 msec −2 .
13. Two bodies are projected vertically upwards from
one point with the same initial velocity v0. The second body is projected t0 second after the first. After
how much time will the bodies meet?
14. A stone is dropped from the top of a tower. When
it crosses a point 5 m below the top, another
stone is dropped from a point 25 m below the
top. Both stones reach the bottom of the tower
simultaneously. Find the height of the tower.
Take g = 10 ms −2 .
15. An elevator without a ceiling is ascending with a
constant speed of 10 ms −1. A boy on the elevator
throws a ball directly upward, from a height of 2 m
04_Kinematics 1_Part 1.indd 48
above the elevator floor, just as the elevator floor is
28 m above the ground. The initial speed of the ball
with respect to the elevator is 20 ms −1.
(a)What maximum height above the ground does
the ball reach?
(b)How long does the ball take to return to the
elevator floor?
Take g = 10 ms −2
16. A ball is projected vertically upwards with a velocity of 24.5 msec −1 from the bottom of a tower.
A boy, who is standing at the top of the tower is
unable to catch the ball when it passes him in the
upward direction. But the ball again reaches him
after 3 second when it is falling and then he catches
it. Find what was the velocity of the ball when the
boy caught it and find also the height of the tower.
17. Two particles begin a free fall from rest from the
same height, 1 s apart. How long after the first particle begins to fall will the two particles be 10 m
apart?
Take g = 10 ms −2 .
11/28/2019 7:06:08 PM
Motion in a Plane and Relative Velocity
MOTION IN A PLANE: AN INTRODUCTION
Any type of planar motion can be resolved into two
rectangular rectilinear motion i.e., two mutually perpendicular independent motions resolved along x
and y axis (since x and y components do not have
any dependence in each other). Consider a particle to
be moving along a curve C in x -y plane. Let it be at
a point P ( x , y ) which has a position vector r at any
particular instant of time t . Let this position vector
make an angle q with position x-axis as shown in
Figure.
vy
ϕ
r
v y = uy + a y t
1
x = ux t + a x t 2
2
1
y = uy t + a y t 2
2
v 2x - u2x = 2a x x
v 2y - u2y = 2a y y
⎛ u + vx ⎞
x=⎜ x
t
⎝ 2 ⎟⎠
⎛ uy + v y ⎞
y=⎜
t
⎝ 2 ⎟⎠
The most significant thing about these types of motion
is that they are independent of each other.
Illustration 49
Calculate the displacement, when a particle is
displaced
From Figure, we get
x = r cos q and y = r sin q
y
Also, r = r 2 + y 2 and tan q =
x
Let v be the velocity of the particle at point P
and let f be the angle made by v with x-axis . So,
again we get
vx = v cos f
(a) 5 m due North and then 12 m due East.
(b) from point 1 to 2 having position vector
r = 2iˆ - 3 ˆj + 4 kˆ m and r = 6iˆ - 6 ˆj + 16 kˆ m.
1
vy
vx
(
2
)
Also find the magnitude in both the cases.
Solution
(a) Dr1 = 5 ˆj
( iˆ )
r
D 2 = 12 ˆ
ˆ
⇒ Dr = Dr1 + Dr2 = 5 j - 12i m
⇒ Dr = 25 + 144 = 13 m
)
N
Δr2
Δr1 = 5j W
Δr
y
N N
S
E
E
SE
A similar and identical treatment can also be done for
the acceleration components ax and ay . Further, the
equations of motion in vector form are
v = u + at
04_Kinematics 1_Part 2.indd 49
)
(
vy = v sin f
1
r = ut + at 2
2
2
2
v - u = 2 a.r
⎛ u+v⎞
r=⎜
t
⎝ 2 ⎟⎠
(
W
x
v = vx2 + vy2 and tan f =
v x = ux + a x t
W N
θ
y-components of
equations of motion
SW
y
v
vx
x-component of
equations of motion
E
x
S
(b) Dr = r2 - r1 = r f - ri
⇒
⇒
(
(
) (
)
Dr = 6iˆ - 6 ˆj + 16 kˆ - 2iˆ - 3 ˆj + 4 kˆ
Dr = 4iˆ - 3 ˆj + 12kˆ m
)
11/28/2019 7:49:19 PM
4.50 JEE Advanced Physics: Mechanics – I
2
2
2
Dr = ( 4 ) + ( 3 ) + ( 12 ) = 13 m
⇒
1
Δr = r2 – r1
r1
2
r2
A particle travels along the parabolic path given by
y = bx 2 . If its component of velocity along the y-axis
is vy = ct 2 , determine the x and y components of
the particle’s acceleration. Assume b and c to be
positive constants.
Solution
Since vy = ct 2
dy
= ct 2
dt
⇒
dy = ct 2 dt
y
⇒
2
⇒
y=
0
c 3
t = bx 2
3
c 32
t
3b
x=
Thus, the x component of the particle’s velocity can
be determined by taking the time derivative of x
⇒
ay = v y =
d ( 2)
ct = 2ct
dt
Illustration 51
Solution
dr d ⎡ ( 3
v=
=
3t - 2t ) iˆ - ( 4 t + t ) ˆj + ( 3t 2 - 2 ) kˆ ⎤⎦
dt dt ⎣
⎡
⎤
⎛ 2
⎞
v = ⎢ ( 9t 2 - 2 ) iˆ - ⎜
+ 1 ⎟ ˆj + ( 6t ) kˆ ⎥ ms -1
⎝ t
⎠
⎣
⎦
⎤
dv d ⎡ ( 2
⎛ 2
⎞
⇒ a=
= ⎢ 9t - 2 ) iˆ - ⎜
+ 1 ⎟ ˆj + ( 6t ) kˆ ⎥ ms -1
⎝ t
⎠
dt dt ⎣
⎦
⎡
-3 2 ˆ
2
ˆ
ˆ
⇒ a = ⎣ ( 18t ) i + t
j + 6 k ⎤⎦ ms
When t = 2 s ,
⇒
dx d ⎡ c 3 2 ⎤ 3 c 1 2
vx = x =
= ⎢
t ⎥=
t
dt dt ⎣ 3b
⎦ 2 3b
Now again to calculate x and y components of
acceleration, we use
ax =
dvy
dvx
= v x and ay =
= v y
dt
dt
04_Kinematics 1_Part 2.indd 50
)
v = ( 9 ( 22 ) - 2 ) iˆ - ( 2 ( 2 -1 2 ) + 1 ) ˆj + 6 ( 2 ) kˆ ms -1
⇒ v = 34iˆ - 2.414 ˆj + 12kˆ ms -1
So, the magnitude of the particle’s velocity is
c 3
t …(1)
3
Substituting y from (1) in y = bx 2, we get
⇒
⇒
(
t
0
d ⎛ 3 c 1 2 ⎞ 3 c -1 2 3 c 1
t ⎟=
t
=
⎜
⎠ 4 3b
4 3b t
dt ⎝ 2 3b
(
∫ dy = ∫ ct dt
ax = v x =
The position of a particle is r = ( 3t 3 - 2t ) iˆ - ( 4 t + t ) ˆj + ( 3t 2 - 2 ) kˆ m
r = ( 3t 3 - 2t ) iˆ - ( 4 t + t ) ˆj + ( 3t 2 - 2 ) kˆ m , where t is in seconds. Determine the
magnitude of the particle’s velocity and acceleration
when t = 2 s .
Illustration 50
⇒
⇒
)
2
v = vx2 + vy2 + vz2 = 34 2 + ( -2.414 ) + 122
⇒
v = 36.1 ms -1 Acceleration
When t = 2 s ,
a = ⎣⎡ 18 ( 2 ) iˆ + 2 -3 2 ˆj + 6 kˆ ⎤⎦ ms -2
⇒ a = ⎡⎣ 36iˆ + 0.3536 ˆj + 6 kˆ ⎤⎦ ms -2
So, the magnitude of the particle’s acceleration is
2
a = ax2 + ay2 + az2 = 36 2 + ( 0.3536 ) + 6 2 = 36.5 ms -2
Illustration 52
The velocity of a particle is v = ⎡⎣ 3iˆ + ( 6 - 2t ) ˆj ⎤⎦ ms -1 ,
where t is in seconds. If r = 0 when t = 0 , determine
the displacement of the particle during the time interval t = 1 s to t = 3 s .
11/28/2019 7:49:27 PM
Chapter 4: Kinematics I
Solution
The position r of the particle can be determined by integrating the kinematic equation
dr
v=
( i.e., dr = vdt ) and then using the initial condt
dition i.e., at t = 0 , r = 0 for solving the integral.
dr = vdt
r
⇒
t
∫ dr = ∫ ⎡⎣ 3iˆ + ( 6 - 2t ) ˆj ⎤⎦ dt
0
⇒
0
r = ⎡⎣ 3tiˆ + ( 6t - t 2 ) ˆj ⎤⎦ m
(
(
)
)
Thus, the displacement of the particle is
Dr = r t = 3 s - r t =1 s
⇒ Dr = 9iˆ + 9 ˆj - 3iˆ + 5 ˆj
ˆ
ˆ
⇒ Dr = 6i + 4 j m
) (
)
3
= 0.9 ms -2
5
At t = 0 , ux = 8 ms -1 , uy = 0
and ay = ( 1.5 ms -2 ) ×
ax = 1.2 ms -2 , ay = 0.9 ms -2
x = 0 and y = 0
The x-component of the velocity at time t = 4 s is
given by
vx = ux + ax t
When t = 1 s and 3 s ,
r t=1 s = 3 ( 1 ) iˆ + [ 6 ( 1 ) - 12 ] ˆj = 3iˆ + 5 ˆj ms -1
r t= 3 s = 3 ( 3 ) iˆ + [ 6 ( 3 ) - 3 2 ] ˆj = 9iˆ + 9 ˆj ms -1
(
(
⇒
vx = 8 ms -1 + ( 1.2 ms -2 ) ( 4 s ) ⇒
vx = 8 ms -1 + 4.8 ms -1 = 12.8 ms -1 The y-component of velocity at t = 4 s is given by
v y = uy + ay t
⇒
A particle is moving in the x -y plane with constant
acceleration of 1.5 ms -2 that makes an angle of 37°
with the x-axis. At t = 0 the particle is at the origin
and its velocity is 8 ms -1 along the x-axis. Find the
velocity and the position of the particle at t = 4 s .
Given sin ( 37° ) = 0.6
vy = 0 + ( 0.9 ms -2 ) ( 4 s ) = 3.6 ms -1
The velocity of the particle at t = 4 s is
)
Illustration 53
v = vx2 + vy2 =
( 12.8 ms -1 )2 + ( 3.6 ms -1 )2
v = 13.3 ms -1
Assume that the velocity makes an angle q with the
x-axis, then
vy
3.6 ms -1
9
tan q =
=
=
-1
vx 12.8 ms
32
⎛ 9 ⎞
⇒ q = tan -1 ⎜ ⎟
⎝ 32 ⎠
The x-coordinate at t = 4 s is given by
y
1 2
ax t
2
1
2
x = ( 8 ) ( 4 ) + ( 1.2 )( 4 )
2
x = 32 m + 9.6 m = 41.6 m x = ux t +
⇒
a = 1.5 ms–2
37°
u = 8 ms–1
⇒
x
The y-coordinate at t = 4 s is given by
y = uy t +
Solution
Let us first calculate the components of constant
acceleration. If ax and ay be the respective acceleration components along x and y -axis then
4
ax = ( 1.5 ms -2 ) ( cos 37° ) = ( 1.5 ms -2 ) × = 1.2 ms -2
5
04_Kinematics 1_Part 2.indd 51
4.51
⇒
⇒
1 2
ay t
2
1
( 0.9 ) ( 4 )2 2
y = 7.2 m y=
{ uy = 0 }
Thus, the coordinates of the particle at 4 s are (41.6 m,
7.2 m).
11/28/2019 7:49:35 PM
4.52 JEE Advanced Physics: Mechanics – I
Illustration 54
A particle is moving in a plane with velocity given by
v = u0 iˆ + [ aw cos ( w t ) ] ˆj . If the particle is at the origin
at t = 0 ,
⎛ 3π ⎞
So, distance from the origin at t = ⎜
will be
⎝ 2w ⎟⎠
2
(a) calculate the trajectory of the particle
⎛ 3π ⎞
.
(b) find its distance from the origin at time ⎜
⎝ 2w ⎟⎠
⇒
⎛ 3π u0 ⎞
r = x2 + y2 = ⎜
+ a2
⎝ 2w ⎟⎠
r = a2 +
Solution
v = vx iˆ + vy ˆj
⇒ vx =
vy =
⇒ dx = u0 dt
⇒
y
⇒
t
∫ dx = u ∫ dt
dy
= aw cos ( w t )
dt
t
∫ dy = aw ∫ cos ( wt ) dt
0
0
0
0
t ⎞
0
⇒ x = u0 t …(1)
⎛ sin ( w t )
⇒ y = aw ⎜
⎟
⎝
w
0⎠
⇒ y = a sin ( w t ) …(2)
(a) From (1), t =
x
. Substituting in (2), we get
u0
⎛ wx ⎞
y = a sin ⎜
⎝ u0 ⎟⎠
{Equation of Trajectory}
This is the desired trajectory and it is a sine curve
as shown in figure.
y
+a
π u0
ω
2π u0
ω
O
A particle of mass 2 kg has a velocity of 2 ms -1 .
A constant force of 4 N acts on the particle for 1 s in a
direction perpendicular to its initial velocity. Find the
velocity and displacement of the particle at t = 1 s .
Solution
Let the velocity of the particle be along x -axis. Then
u = 2iˆ
So, force on the particle is
F = 4 ˆj
F
⇒ a = = 2 ˆj
m
Since a = constant, so we can use
1
v = u + at and s = ut + at 2
2
v = u + at
⇒ v = 2iˆ + ( 2t ) ˆj ⇒ v t=1 s = 2iˆ + 2 ˆj So, the situation is shown in Figure (a).
at = 2 ms–1
x
v
–a
⎛ 3π ⎞
(b) For t = ⎜
from equations (1) and (2), we get
⎝ 2w ⎟⎠
⎛ 3π ⎞
x = u0 ⎜
and y = - a
⎝ 2w ⎟⎠
04_Kinematics 1_Part 2.indd 52
4w 2 Illustration 55
dx
= u0
dt
x
9π 2 u02
α
u = 2 ms–1
Figure (a)
If v makes an angle a with x-axis, then
at
⎛ 2⎞
a = tan -1 = tan -1 ⎜ ⎟ = tan -1 ( 1 ) = 45°
⎝ 2⎠
u
11/28/2019 7:49:40 PM
Chapter 4: Kinematics I
Thus, velocity of the particle at the end of 1 s is
2 2 ms -1 at an angle of 45° with its initial velocity.
The situation is again shown in Figure (b).
1 2
at = 1 m
2
sOLUTION
(a) ux = 3 ms -1
ax = -1 ms -2 and ay = -0.5 ms -2
1 2
ax t
2
For x to be MAXIMUM, we have
Since x = ux t +
s
β
4.53
ut = 2 m
⇒
dx
=0
dt
1
ux + ax ( 2t ) = 0
2
ux + ax t = 0
⇒
t=t=-
Figure (b)
1
Since s = ut + at 2
2
⇒
…(1)
⇒
1
s = ( 2t ) iˆ + ( 2 ) t 2 ˆj
2
s
= 2iˆ + ˆj
⇒
s = 5m
⇒
⎛ 1 2 ⎞
at
⎟
⎛ 1⎞
b = tan ⎜ 2 ⎟ = tan -1 ⎜ ⎟
⎝ 2⎠
⎝ ut ⎠
Please note that this happens to be the time when
the x motion of the particle is reversed i.e., this
time is the time when the x component of velocity should be zero and equation (1) shows that.
At this instant
⇒
t= 1 s
If b is the angle which s makes with x-axis, then
-1 ⎜
So, displacement of the particle at the end of 1 s
⎛ 1⎞
is 5 m at an angle of tan -1 ⎜ ⎟ from its initial
⎝ 2⎠
velocity.
ILLUsTRATION 56
A particle leaves the origin with an initial velocity
v = ( 3iˆ ) ms -1
and
a
constant
acceleration
2
a = -1iˆ - 0.5 ˆj ms . When the particle reaches its
(
)
maximum x coordinate, what are
3
( -1 )
=3s
vx = 0
and vy = uy + ay t = 0 - 0.5 × 3 = -1.5 ms -1
(
)
⇒ iˆ v = -1.5 ˆj ms -1
(b) x = ux t +
1 2
1
2
ax t = 3 × 3 + ( -1 ) ( 3 ) = 4.5 m
2
2
1 2
1
2
ay t = 0 - ( 0.5 )( 3 ) = -2.25 m
2
2
⇒ iˆ r = 4.5iˆ - 2.25 ˆj m
y = uy t +
(
(a) its velocity and
(b) its position vector?
ux
ax
)
Test Your Concepts-VI
Based on Planar Motion
(Solutions on page H.85)
1. A particle starts from the origin at t = 0 with a
velocity of 8 ĵ ms -1 and moves in the xy plane with
a constant acceleration of ( 4iˆ + 2 ˆj ) ms -2 . At the
instant the particle’s x coordinate is 32 m, what are
(a) its y-coordinate and
(b) its speed?
04_Kinematics 1_Part 2.indd 53
2. A particle starts from the origin of the coordinates
along the path defined by the parabola y = 0.5 x 2 .
If the component of velocity along the x-axis is
v x = ( 5t ) ms -1, where t is in second, determine
the particle’s distance from the origin O and the
magnitude of acceleration when t = 1 s.
11/28/2019 7:49:46 PM
4.54 JEE Advanced Physics: Mechanics – I
y
y = 0.5 x 2
O
y
x
3. A point moves the plane x-y according to the law
x = k sin( w t ) and y = k [ 1- cos ( w t ) ] where k and
w are positive constants. Find the distance traversed by the particle during time t.
4. A particle moves in the plane according to the law
x = kt, y = kt ( 1- a t ), when k and a are positive
constants, and t is the time. Find
(a) the equation of the particle’s trajectory y(x)
(b)the velocity v and the acceleration a of the
point as a function of time.
5. The speed of a particle moving in a plane is equal
to the magnitude of its instantaneous velocity,
v = v = v 2x + v 2y . Show that the rate of change of
dv a ⋅ v
= .
the speed is
dt
v
6. As soon as a rocket reaches an altitude of 40 m
it begins to travel along the parabolic path
( y - 40 )2 = 160 x, where the coordinates are
RELATIVE MOTION
Motion is a combined property of the object under
study as well as the observer. It is always relative
because there is no such thing like absolute motion or
absolute rest. Motion is always defined with respect to an
observer or a reference frame.
Reference Frame
The motion of a particle is described by using kinematical quantities such as velocity, acceleration.
However, these quantities are dependent upon the
state of motion as seen by the observer. So, to understand the concept of relative motion it becomes mandatory for us to talk about or introduce the concept of
a “reference frame”.
04_Kinematics 1_Part 2.indd 54
measured in meters. Assuming that the component of velocity in the vertical direction is constant
at v y = 180 ms -1, determine the magnitudes of the
rocket’s velocity and acceleration when it reaches
an altitude of 80 m.
( y – 40)2 = 160x
40 m
x
7. Velocity and acceleration of a particle at time t = 0
are u = ( 2iˆ + 3 ˆj ) ms -1 and a = ( 4iˆ + 2 ˆj ) ms -2
respectively. Find the velocity and displacement of
particle at t = 2 s.
8. A balloon is ascending at the rate v = 12 kmh-1
and is being carried horizontally by the wind blowing at v w = 20 kmh-1. If a ballast bag is dropped
from the balloon at the instant h = 50 m, determine the time required for it to strike the ground.
Assume that the bag was released from the balloon with the same velocity as the balloon. Also,
with what speed does the bag strike the ground?
Take g = 10 ms -2 .
(
)
Reference frame is an axis system from which
motion is observed along with a clock attached to the
axis, to measure time. Reference frame can be stationary or moving.
In layman language a Reference Frame is a platform from where the observer observes motions and
takes measurements with respect to it.
Suppose there are two persons A and B sitting
in a car moving at constant speed. Two stationary
persons C and D observe them from the ground.
D
C
B A
Here B appears to be moving for C and D, but at rest
for A. Similarly C appears to be at rest for D but
moving backward for A and B .
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Chapter 4: Kinematics I
Relative Motion In One Dimension
Relative Position
It is the position of a particle w.r.t. observer. In general if position of A w.r.t. origin is x A and that of
B w.r.t. origin is xB , then position of A w.r.t B (i.e.
B is now the new origin) is denoted by x AB and is
given by
x AB = x A - xB
xB
xAB
B
and vAB = velocity of A w.r.t. B =
⇒
vAB =
A
dx AB
d
= ( x A - xB )
dt
dt
dx A dxB
= vA - vB
dt
dt
So, for particles A and B moving in the same
direction
vAB = vA - vB
and for particles A and B moving in the opposite
direction
vAB = vA + vB
Relative Velocity
Relative velocity of a particle A with respect to B
is defined as the velocity with which A appears to
move if B is considered to be at rest. In other words,
it is the velocity with which A appears to move as
seen by B considering itself to be at rest.
Conceptual Note(s)
(a) All velocities are relative and have no significance
unless observer is specified. However, when we
say velocity of A, what we mean is, velocity of A
w.r.t. ground which is assumed to be at rest.
(b) Velocity of an object w.r.t. itself is always zero.
(c) Velocity of A and B must always be measured
from the same reference.
04_Kinematics 1_Part 2.indd 55
Illustration 57
Two trains, each having a speed of 30 kmhr -1 are
headed towards each other on the same straight
track. A bird that can fly at 60 kmhr -1 flies off one
train, when they are 60 km apart and heads directly
for the other train. On reaching the other train, it flies
directly back to the first and so on.
(a) How many trips can the bird make from one
train to the other before they crash?
(b) What is the total distance the bird travels?
Solution
xA
Origin
4.55
The relative velocity of one train relative to the other
is 60 kmhr -1 and as the distance between the trains is
60 km, the two trains will crash after 1 hr.
(a) Now, the velocity of bird with respect to train
towards which it is moving will be v = 90 kmhr -1 .
So, the time taken by bird for first trip is
⎛ 60 ⎞ 2
t1 = ⎜ ⎟ = hr and in this time the trains have
⎝ 90 ⎠ 3
⎛ 2⎞
moved towards each other ⎜ ⎟ × 60 = 40 km , so
⎝ 3⎠
the remaining distance = 60 - 40 = 20 km .
So, the time taken by bird for second trip,
20 ⎛ 2 ⎞
= ⎜ ⎟ hr
t2 =
90 ⎝ 3 2 ⎠
Proceeding in the same way time taken by the
⎛ 2 ⎞
bird for nth trip, tn = ⎜ n ⎟ hr
⎝3 ⎠
Now, if the bird makes n trips till the train
crashes,
t1 + t2 + t3 + .... + tn = 1 hr
⇒
2 2
2
+
+ .... + n = 1 hr
3 32
3
⇒
2⎡
1 1
1 ⎤
1 + + 2 + .... + n -1 ⎥ = 1 hr
3 ⎢⎣
3 3
3
⎦
⇒
n
⎡
⎛ 1⎞ ⎤
⎢ 1 - ⎜⎝ ⎟⎠ ⎥
2⎢
3 ⎥
= 1 hr
3⎢
⎛ 1⎞ ⎥
⎢⎣ 1 - ⎜⎝ 3 ⎟⎠ ⎥⎦
⇒
⎛ 1⎞
1- ⎜ ⎟ = 1
⎝ 3⎠
n
11/28/2019 7:49:55 PM
4.56 JEE Advanced Physics: Mechanics – I
⇒
n
⎛ 1⎞
⎜⎝ ⎟⎠ = 0
3
⇒ n → ∞
So, the bird will make infinite trips.
(b) As the speed of bird is 60 kmhr -1 and the two
trains crash after 1 hr., so the total distance travelled by the bird is the distance travelled by the
bird in 1 hr. i.e.,
Illustration 58
When two particles A and B are at point O , A is
moving with a constant velocity 50 ms -1 , while B is
not moving. But B possesses a constant acceleration
of 10 ms -2 . After how much time they will be at a
distance of 125 m ?
Solution
For particle A
uA = 50 ms -1
⎛ km ⎞
d = 60 ⎜
× 1 hr = 60 km
⎝ hr ⎟⎠
aA = 0 ms -2
Relative Acceleration
A
B
It is the rate at which relative velocity is changing
aAB =
dvAB dvA dvB
=
= aA - aB
dt
dt
dt
O
For particle B
Equations of motion when relative acceleration arel is constant are
vrel = urel + arel t
srel = urel t +
1
arel t 2
2
2
2
vrel
= urel
+ 2 arel srel
where urel is the initial relative velocity, vrel is the
final relative velocity i.e. relative velocity at some
instant of time t and srel is the relative displacement
of the particles.
Equations of Motion in Relative
Velocity Form
The kinematical equations of motion are also modified as follows
vr = ur + ar t
1
sr = ur t + ar t 2
2
vr 2 - ur 2 = 2 ar sr
where ur is the initial relative velocity
vr is the final relative velocity at time t
ar is the relative constant acceleration
sr is the relative separation at time t
04_Kinematics 1_Part 2.indd 56
uB = 0
aB = 10 ms -2
So, initial velocity of A w.r.t. B
uAB = uA - uB = 50 ms -1
and acceleration of A w.r.t. B
aAB = aA - aB = -10 ms -2
The distance between A and B after time t is given
by
sAB = uAB t +
1
aAB t 2
2
⇒ 125 = 50t - 5t 2 ⇒
t 2 - 10t + 25 = 0 ⇒
( t - 5 )2 = 0 ⇒
t= 5 s
Illustration 59
Car A and car B start moving simultaneously in
the same direction along the line joining them. Car
A with a constant acceleration a = 4 ms -2 , while car
B moves with a constant velocity v = 1 ms -1 . At time
t = 0 , car A is 10 m behind car B . Find the time
when car A overtakes car B .
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Chapter 4: Kinematics I
Relative Motion In Two Dimension
Solution
-1
Given uA = 0 , uB = 1 ms , aA = 4 ms
a = 4 ms–2
A
-2
and aB = 0
v = 1 ms–1
B
10 m
Let rA be the position of A with respect to O i.e.
position vector of the point A is rA .
Similarly let rB be the position of B with respect to
O i.e. position vector of the point B is rB .
+ve
y
A
Assuming car B to be at rest, we have
rAB
uAB = ur = uA - uB = 0 - 1 = -1 ms -1
aAB = ar = aA - aB = 4 - 0 = 4 ms
10 m
aAB = 4 ms–2
B
at rest
Substituting the proper values in equation
1 2
ar t
2
1
2
⇒ 10 = -t + ( 4 ) ( t )
2
sr = ur t +
⇒
⇒
⇒
⇒
rA
B
-2
Now, the problem can be assumed in simplified form
as follows:
uAB = –1 ms–1
A
4.57
2t 2 - t - 10 = 0 1 ± 1 + 80 1 ± 81
t=
=
4
4
1± 9
t=
4 t = 2.5 s and -2 s Ignoring the negative value, the desired time is 2.5 s
rB
x
O
Then the position of A with respect to B is rAB
given by
rAB = rA - rB
(The vector sum rA - rB can be done by D law of
addition or resolution method).
Now the velocity of A with respect to B is vAB ,
which is the rate at which position of A with respect
to B changes i.e.
d vAB = ( rAB )
dt
dr
dr
d d d ( rAB ) = ( rA ) - ( rB ) = A - B = vA - vB
dt
dt
dt
dt
dt
⇒ vAB = vA - vB So, vAB = vA B = vA - vB = velocity of A relative to B
⇒
Similarly
vBA = vB A = vB - vA = velocity of B relative to A
Problem Solving Technique(s)
This problem can also be solved without using the
concept of relative motion as under,
At the time when A overtakes B,
s A = sB + 10
⇒
1
× 4 × t 2 = 1× t + 10
2
⇒
2t 2 - t - 10 = 0
Which on solving gives t = 2.5 second and -2 second,
the same as we found above.
04_Kinematics 1_Part 2.indd 57
vB
vBA
θ
vB
vAB
θ
vA
vA
If vA and vB are inclined to each other at angle q ,
then
vAB = vBA = vA 2 + vB 2 - 2v2 vB cos q
and vAB = -vBA 11/28/2019 7:50:11 PM
4.58 JEE Advanced Physics: Mechanics – I
Problem Solving Technique(s)
(a) Since v AB = v A - vB
⇒ v AB = v A - vB + vP - vP
⇒ v AB = ( v A - vP ) - ( vB - vP ) = v AP - vBP
⇒ v AB = v A - vB = v AP - vBP = v AO - vBO
i.e., relative velocity is simplify independent of the
velocity of the observer (O) analysing the motion,
as long as both the particles A and B have their
velocities specified w.r.t. the same observer.
By default v A and vB are specified with respect
to the ground or with respect to a stationary
observer at the ground.
(b) If Dr is the relative separation between any two
particles at time Dt, then
Dr = vr Dt
2
2
⇒ Dr = v A + vB - 2v A vB cosq Dt
(
)
(c) A similar treatment is also given when we are
asked to deal with the relative acceleration a AB
or aBA
a AB = a A - aB = Acceleration of A relative to B
a
a
BA = B - a A = Acceleration of B relative to A
Similarly
2
2
a AB = aBA = a A + aB - 2a A aB cosq
But here we must keep one thing in mind that
if we are asked to calculate the acceleration of
a ball falling freely with respect to another ball
projected vertically upwards, then relative acceleration is zero (and not 2g). This is due to the fact
that both the balls (either the ball falling freely or
the one launched upward) move under the influence of gravity which acts vertically downwards
for both.
Similarly, the acceleration of A with respect to B is
aAB , which is the rate at which velocity of A with
respect to B changes i.e.
d aAB = ( vAB )
dt
dvA dvB d d d ⇒
= aA - aB
( vAB ) = ( vA ) - ( vB ) =
dt
dt
dt
dt
dt
⇒ aAB = aA - aB 04_Kinematics 1_Part 2.indd 58
If aAB is acceleration of A relative to B and aBA is
acceleration of B relative to A , then
aAB = aA - aB and aBA = aB - aA
⇒ aAB = - aBA Relative Motion For Bodies Moving
Independently
For two bodies moving independently in the same
direction, relative velocity is vr = vA ∼ vB = vA - vB
(~ sign indicates positive difference between vA and
vB )
This relative velocity can be the relative velocity of approach or of separation depending upon the
location of the bodies and the magnitudes of their
velocities.
vr = vA - vB , is the relative velocity of approach if
vA > vB and A is following (or approaching) B
A
B
Relative separation between A and B decreases with time
Whereas the same vr becomes relative velocity of
separation when B is behind A .
B
vB
vA
A
Relative separation between A and B increases with time
For two bodies moving independently in opposite
direction
vr = vA + vB
is the relative velocity of approach till the bodies do
not meet and after they meet, the same value is the
relative velocity of separation (or receding).
A
B
Bodies approaching and relative separation decreasing with time
B
A
Bodies receding and relative separation increasing with time
Relative Motion For Bodies Moving
Dependently
Dependent motion is the case when a person is moving on a conveyer belt or on the surface of a moving
carriage train or an escalator.
11/28/2019 7:50:16 PM
Chapter 4: Kinematics I
In such cases, we have to first understand that “if
a velocity of the person walking on a moving surface is
given, then this velocity is the velocity of the person with
respect to the surface (i.e. his ground) on which he is moving and not with respect to the stationary ground.”
Let’s discuss and understand this through the
­following situations.
Situation 1
Consider a train T moving with a velocity vT . Let a
person P move with a velocity v (say) on this train
along the direction of the train, then the velocity of
the person with respect to his ground (which is the
train) is v , so
vPT = v , where vPT = vP - vT
⇒
vP = vT + vPT which simply means that the velocity of the person
with respect to a stationary observer A at the ground
is
vP = vT + v
Situation 1a
If the observer A also starts moving on the ground in
the direction of motion of the train, then he sees the
person to be moving with a velocity
vPA = ( vT + v ) - vA
Situation 1b
If the observer A also starts moving on the ground in
the direction opposite to the motion of the train, then
he sees the person to be moving with a velocity
vPA = ( vT + v ) + vA
4.59
which simply means that the velocity of the person
with respect to a stationary observer at the ground is
vP = vT - v
Situation 2a
If the observer A also starts moving on the ground in
the direction of motion of the train, then he sees the
person to be moving with a velocity
vPA = ( vT - v ) - vA
Situation 2b
If the observer A also starts moving on the ground in
the direction opposite to the motion of the train, then
he sees the person to be moving with a velocity
vPA = ( vT - v ) + vA
Situation 3
If a person ( P ) is moving with a velocity vP on a
stationary escalator ( E ) , then with respect to the
ground ( G ) , we have
vPG = vPE = vP
Situation 3a
If a person ( P ) is moving with a velocity v up an
escalator ( E ) which is also moving up with a velocity vE , then with respect to the ground ( G ) , we have
vPG = vPE + vEG
Since, we must take a note here that vPE = v , so velocity of the person with respect to ground or a stationary observer A on the ground is
vP = vE + v
Situation 2
Situation 3b
Similarly if the person P moves with a velocity v (say)
on this train opposite to the direction of the train, then
the velocity of the person with respect to his ground
(which is the train) is v , so
If a person ( P ) is moving with a velocity v down an
escalator ( E ) which is moving up with a velocity vE ,
then with respect to the ground ( G ) , we have
vPT = -v , where vPT = vP - vT
(The direction of motion of train is taken as positive)
⇒
vP = vT + vPT
vPG = vPE + vEG
Taking direction of motion of the escalator as positive, we note here that vPE = -v and vEG = + vE , so
velocity of the person with respect to ground or a stationary observer A on the ground is
vP = vE - v
04_Kinematics 1_Part 2.indd 59
11/28/2019 7:50:21 PM
4.60 JEE Advanced Physics: Mechanics – I
2π
Illustration 60
Consider a collection of a large number of particles
each with speed v . The direction of velocity is randomly distributed in the collection. Show that the
magnitude of the relative velocity between a pair of
particles averaged over all the pairs in the collection
is greater that v .
⎡
⎛q⎞⎤
-4v ⎢ cos ⎜ ⎟ ⎥
⎝ 2⎠ ⎦
⎣
2v
0
=
= - ( cos π - cos 0 )
2π - 0
π
⇒
v21
⇒
v21 = -
⇒
v21 > v
2v
4v
( -1 - 1 ) =
= 1.273v
π
π
Solution
As shown in figure, let v1 and v2 be the velocities of
any two particles and q be the angle between them.
As each particle has speed v , so
v1 = v2 = v
The magnitude of relative velocity v21 of particle 2
with respect to 1 is given by
v21 =
⇒
2 2
-v1 + v2 + 2 -v1 v2 cos ( 180° - q )
v21 = v 2 + v 2 - 2 ( v )( v ) cos q = 2v 2 ( 1 - cos q ) ⎛q⎞
Since 1 - cos q = 2 sin 2 ⎜ ⎟
⎝ 2⎠
⇒
As the velocities of the particles are randomly distributed, so q can vary from 0 to 2π . If v21 is the magnitude of the average velocity when averaged over
all pairs, then
2π
⎛q⎞
∫ 2v sin ⎜⎝ 2 ⎟⎠ dq
v21 = 0
2π
=
⎡
⎛q⎞ ⎤
⎢ cos ⎜⎝ 2 ⎟⎠ ⎥
-2v ⎢
⎥
⎢ ⎛⎜ 1 ⎞⎟ ⎥
⎢⎣ ⎝ 2 ⎠ ⎥⎦ 0
∫ dq
q
2π
0
0
C
B
v21
v2
180° – θ
A′
04_Kinematics 1_Part 2.indd 60
–v1
θ
O
v1
Two trains, each having a speed of 30 kmhr -1 are
headed towards each other on the same straight
track. A bird that can fly at 60 kmhr -1 flies off one
train, when they are 60 km apart and heads directly
for the other train. On reaching the other train, it flies
directly back to the first and so on.
(a) How many trips can the bird make from one
train to the other before they crash?
(b) What is the total distance the bird travels?
Solution
The relative velocity of one train relative to the other
is 60 kmhr -1 and as the distance between the trains is
60 km, the two trains will crash after 1 hr.
⎛q⎞
⎛q⎞
v21 = 2v 2 × 2 sin 2 ⎜ ⎟ = 2v sin ⎜ ⎟
⎝ 2⎠
⎝ 2⎠
2π
Illustration 61
A
(a) Now, the velocity of bird with respect to train
towards which it is moving will be v = 90 kmhr -1 .
So, the time taken by bird for first trip is
⎛ 60 ⎞ 2
t1 = ⎜ ⎟ = hr
⎝ 90 ⎠ 3
and in this time the trains have moved towards
⎛ 2⎞
each other ⎜ ⎟ × 60 = 40 km , so the remaining
⎝ 3⎠
distance = 60 - 40 = 20 km .
So, the time taken by bird for second trip,
20 ⎛ 2 ⎞
= ⎜ ⎟ hr
t2 =
90 ⎝ 3 2 ⎠
Proceeding in the same way time taken by the
⎛ 2 ⎞
bird for nth trip, tn = ⎜ n ⎟ hr
⎝3 ⎠
Now, if the bird makes n trips till the train
crashes,
t1 + t2 + t3 + .... + tn = 1 hr
⇒
2 2
2
+ 2 + .... + n = 1 hr
3 3
3
11/28/2019 7:50:26 PM
Chapter 4: Kinematics I
⇒
2⎡
1 1
1 ⎤
1 + + 2 + .... + n -1 ⎥ = 1 hr
⎢
3⎣
3 3
3
⎦
⇒
n
⎡
⎛ 1⎞ ⎤
⎢ 1 - ⎜⎝ ⎟⎠ ⎥
2⎢
3 ⎥
= 1 hr
3⎢
⎛ 1⎞ ⎥
1
⎜
⎟
⎢⎣
⎝ 3 ⎠ ⎥⎦
⇒
⎛ 1⎞
1- ⎜ ⎟ = 1
⎝ 3⎠
⇒
(b) Find the speed of the ball with respect to the
4q
­surface, if f =
.
3
Solution
(a) Since the motion of the ball ( say B ) is observed
from the frame of the trolley A , so in this prob
lem we must deal with vBA i.e., velocity of ball B
with respect to the trolley A .
n
y
n
⎛ 1⎞
⎜⎝ ⎟⎠ = 0
3
A(at t)
⇒ n → ∞ So, the bird will make infinite trips.
v0t
(b) As the speed of bird is 60 kmhr -1 and the two
trains crash after 1 hr., so the total distance travelled by the bird is the distance travelled by the
bird in 1 hr. i.e.,
4.61
A(at t = 0)
vB
⎛ km ⎞
d = 60 ⎜
× 1 hr = 60 km
⎝ hr ⎟⎠
ϕ
On a frictionless horizontal surface, assumed to
be x -y plane, a small trolley A is moving along a
straight line parallel to the y-axis (see figure) with
a constant velocity of ( 3 - 1 ) ms -1 . At a particular
instant, when the line OA makes an angle of 45°
with the x-axis, a ball is thrown along the surface
from the origin O . Its velocity makes an angle f with
the x-axis and it hits the trolley.
x
x0
Now vBA = vB - vA …(1)
Since f is the angle made by the velocity of the
ball with the x-axis, to hit the trolley, so
vB = vB cos f iˆ + sin f ˆj …(2)
Also, q is the angle made by the velocity vector
of the ball with the x-axis in the frame of the trolley B , so
iˆ vBA = vBA cos q iˆ + sin q ˆj …(3)
Also, we are given that v = v ˆj = ( 3 - 1 ) ˆj ms -1
(
iˆ
)
(
y
A
Substituting all in equation (1), we get
45°
)
A
A
x
(a) The motion of the ball is observed from the frame
of the trolley. Calculate the angle q made by the
velocity vector of the ball with the x-axis in this
frame.
04_Kinematics 1_Part 2.indd 61
45°
O
Illustration 62
O
y0
⇒
(
)
iˆ vBA cos q iˆ + sin q ˆj = ( vB cos f ) iˆ +
iˆ ( v sin f - vA ) ˆj …(4)
B
vBA cos q = vB cos f …(5)
and vBA sin q = vB sin f - vA …(6)
So, tan q =
vB sin f - vA
…(7)
vB cos f
11/28/2019 7:50:31 PM
4.62 JEE Advanced Physics: Mechanics – I
Now, let the ball B strike the trolley A at time t,
then the y coordinate of the ball as well as the
trolley have to be the same, so
⇒
( y )Ball at t = ( y )Trolley at time t
Illustration 63
Particle A is at rest and particle B is moving with
constant velocity v as shown in the figure at t = 0 .
Find their velocity of separation.
( vB sin f ) t = y0 + v0 t …(8)
( vB cos f ) t = x0 …(9)
Since we know observe that x0 = y0 , so from (8)
and (9), we get
⇒
( vB sin f ) t = ( vB cos f ) t + v0 t
Solution
vB sin f - v0 = vB cos f …(10)
tan q = 1
vBA = vB - vA = v
vsep = component of vBA along line AB = v cos q
Illustration 64
q = 45°
Also, we could have arrived the result without these
calculations. Just keep in mind that for the ball (B) to
hit the trolley A, relative velocity of the ball B with
respect to the trolley A must be directed along OA.
So, vBA makes an angle of 45° with the x-axis. Hence
q = 45°.
Particles A and B are moving as shown in the figure
at t = 0 . Find their velocity of separation
(a) at t = 0
(b) at t = 1 s
4 ms–1
A
4q 4 ( 45° )
(b) Since f =
=
= 60°
3
3
3m
So, from (7), we get
tan 45° =
vB sin ( 60° ) - ( 3 - 1 )
vB cos ( 60° )
⎛ 3⎞
v
- 3 +1= B
vB ⎜
⎝ 2 ⎟⎠
2
⇒
⎛ 3 -1⎞
vB ⎜
= 3 -1
⎝ 2 ⎟⎠
⇒
vB = 2 ms -1
Velocity of Approach/Separation In
Two Dimension
Relative velocity of approach of two bodies is the relative velocity of the bodies along the line joining the
two bodies. If the separation is decreasing, we say it
is velocity of approach and if separation is increasing,
then we say it is velocity of separation.
04_Kinematics 1_Part 2.indd 62
v
A
Using (10) and (7), we get
⇒
θ
B
Also, we observe that
3 ms–1
B
4m
Solution
(a) tan q =
3
4
vsep = relative velocity along line AB
4 ms–1
4 cos θ
4 sin θ
A
3 sin θ
3m
θ
4m
B
3 ms–1
3 cos θ
11/28/2019 7:50:35 PM
Chapter 4: Kinematics I
⇒
vsep = 3 cos q + 4 sin q
⇒
⎛ 4⎞
⎛ 3 ⎞ 24
vsep = 3 ⎜ ⎟ + 4 ⎜ ⎟ =
= 4.8 ms -1
⎝ 5⎠
⎝ 5⎠
5
(b) q = 45°
In time dt, the person travels a distance ds = vdt
…(1)
Further in DPQM ,
⇒
( ds )2 = ( dr )2 + ( rdq )2 …(2)
where, dr = distance moved by particle along radial
direction rdq = distance moved by particle along tan
gential direction (normal to r )
4 ms–1
4 cos θ
4 sin θ
L
A
M
ro
3 sin θ
7m
θ = 45°
7m
a
3 ms–1
B
3 cos θ
⇒
⇒
vsep = relative velocity along line AB
vsep = 3 cos q + 4 sin q
7
⎛ 1 ⎞
⎛ 1 ⎞
vsep = 3 ⎜
ms -1
+ 4⎜
=
⎝ 2 ⎟⎠
⎝ 2 ⎟⎠
2
Four persons K , L , M and N are initially at the
four corners of a square of side a . Each person now
moves with a uniform speed v in such a way that K
moves always directly towards L , L directly towards
M , M directly towards N and N directly towards
K . Find the path-equation followed by the persons.
Also calculate the time taken by them to meet.
Solution
The symmetry of configuration implies that each person follows similar trajectories and perform similar
motion. They, therefore meet at the same time at the
centre O of the square. Since they are always directing their velocities to the next neighbours, therefore
they must follow a curved path as shown in the figure. The persons are always located at the corners of
the square which rotates and shrinks in size and ultimately collapses to zero.
Path Equation Followed by Persons
Let r be the position vector of point M at any instant
t . The velocity v of the person M always makes
fixed angle φ = 45° with r
v
r
O
K
N
Further in DPQM ,
rdq
dr ⇒
tan 45° =
⇒
dr = rdq …(3)
r
Illustration 65
04_Kinematics 1_Part 2.indd 63
4.63
⇒
∫
a
2
dr
=
r
q
∫ dq
0
r
⇒ q = log e r a
2 ⇒
r×
⇒
r=
2
= eq
a
a
2
eq {Equation of Trajectory}
L
a
a
a
K
M
a
N
where q is the angle measured from M along the
line OM . Putting (3) in (2), generates
⇒
( ds )2 = 2 ( dr )2 11/28/2019 7:50:42 PM
4.64 JEE Advanced Physics: Mechanics – I
⇒
ds = 2dr ⇒
ds
dr
= 2
dt
dt ⇒
dr
v= 2
dt \
r
⇒
⇒
2
dθ
a
2
T
0
0
∫ dr = v∫ dt
rdθ
r
M
Conceptual Note(s)
The above analysis can be applied to any other configuration where moving object makes a fixed angle
f with r
q
tanf
Then equation of trajectory will be r = ro e
r
Time taken by the objects to meet T = o
v cos f
D
a
a
a
C
E
a
a
B
A
a
B
F
a
a
A
e.g., For equilateral triangle f = 30°, ro =
04_Kinematics 1_Part 2.indd 64
a
3
T=
2a
a
=
v cos60 v
⇒
T=
⇒
T=
⇒
T=
Initial Separation
Relative Velocity of Approach
a
⎛ 2π ⎞
v - v cos ⎜
⎝ n ⎟⎠
a
⎡
⎛ 2π ⎞ ⎤
v ⎢ 1- cos ⎜
⎝ n ⎟⎠ ⎥⎦
⎣
a
⎛π⎞
2v sin2 ⎜ ⎟
⎝ n⎠
For triangle n = 3 ⇒
For square, n = 4 ⇒
For hexagon, n = 6
where, T is the time in which object moves from original radial distance r0 to r = 0, with a uniform speed v.
C
and T =
If the particles are located at the sides of an n sided
symmetrical polygon with each side a and each particle moves towards the other, then
Q
a
v
a
2a
a/ 3
=
v cos 30 3v
Shortcut Method to Find Only the
Time Taken
45°
v ds
⎛ a
⎞
- 0⎟ = v(T - 0 )
2⎜
⎝ 2
⎠
T=
3
e 3q and T =
q
r = ae 3
dr
⇒
a
e.g. For hexagon, f = 60°, r0 = a
P
O
r=
⇒
2a
3v
a
T=
v
2a
T=
v
T=
Where to Apply the Concept of
Relative Motion?
The concept of relative motion can be applied to
two and three dimensional motion also. The problems involving the concept of relative motion mainly
belong to the following categories.
(a) Distance of closest approach (i.e., minimum
­distance) between two moving ­bodies
(b) River-Boat problems or River-Swimmer problems
(c) Aeroplane-Wind problems
(d) Rain-Man problems
(e) Relative motion in the case of projectiles
(­discussed in projectile motion)
11/28/2019 7:50:46 PM
Chapter 4: Kinematics I
between the ships and the time after which they are
closest to each other.
A
20 kmh–1
METHOD I (Calculus Method)
dr
=0
dt
So we find or calculate the time at which r attains a
minimum value. Then this value of t is put in r to
get rmin .
METHOD II (Relative Velocity Method)
This method can be applied using the following
steps.
STEP-1
First calculate vBA i.e. velocity of B as seen by A (i.e.
velocity of B as seen by A , when A considers itself
to be at rest).
You can also calculate vAB i.e. velocity of A as
seen by B (i.e. velocity of A seen by B , when B
considers itself to be at rest).
STEP-2
Find the direction of vBA and then draw a line along
the direction of vBA .
STEP-3
Then perpendicular distance between A (consider
ing it to be at rest) and the line along which vBA is
directed is the distance of closest approach or the minimum separation between the two moving bodies.
For this see Illustrative Example below.
Illustration 66
Ship A is moving towards south with a speed of
20 kmh -1 . Another ship B is moving towards east
with a speed of 20 kmh -1 . At a certain instant the
ship B is due south of ship A and is at a distance
of 10 km from ship A . Find the shortest distance
04_Kinematics 1_Part 2.indd 65
y
W
B
N N
E
W N
W
AB = 10 km
E
E
S
SE
For calculating the distance of closest approach (i.e.,
the minimum distance between two moving bodies,
we first calculate the distance r between them at any
instant of time t in the light of some conditions given
with the problem. Then we have to minimise r for
calculating its minimum value. Mathematically, for r
to be MINIMUM, we have
N
SW
CATEGORY 1: DISTANCE OF CLOSEST
APPROACH BETWEEN TWO MOVING
BODIES
4.65
x
S
20 kmh–1
Solution
If we analyse the positions of ships at different
instants, we find that the separation decreases at first,
becomes minimum at a particular instant of time and
then increases.
METHOD I: Using the Concept of Relative Velocity
Attach your reference frame with ship A and then
analyse the motion of ship B . For you the ship A
will be at rest and the ship B will appear to be mov
ing with relative velocity v = 20iˆ + 20 ˆj kmhr -1
(
BA
along line BC .
)
N
y
C
A
Lmin
x
d
P
45° vBA
θ = 45°
B
S
Motion of A relative to B
E
For A , B will appear to be moving with relative
velocity vBA of magnitude 20 2 kmhr -1 at an angle
of 45° with east (i.e., +x axis). There is a point P
on the line of relative motion of B where the separation AB ( = Lmin ) becomes perpendicular to the line
of relative motion. At this instant the separation is
minimum.
vBA
θ
vA
vB
vBA = vB – vA
x
11/28/2019 7:50:52 PM
4.66 JEE Advanced Physics: Mechanics – I
For the triangle ABP,
Lmin
d
sin 45° =
⇒
d
Lmin =
1
hr = 15 min
4
Now the minimum separation as given by equation (1) is
⇒
2
10
=
2
= 5 2 km
2
1⎞
1
⎛
Lmin = ⎜ 10 - 20 × ⎟ + 20 2 × 2
⎝
⎠
4
4
And the time taken is given by
vBA =
2 2
vA + vB = 20 2 kmh -1
1
10 ×
BP
d cos 45°
2 = 1 hr = 15 min
t= =
=
vBA
4
20 2
20 2
v
tan q = A = 1
vB
⇒ q = 45° METHOD II: Using Calculus
We have to find the time after which the separation is
minimum, so let us write separation as a function of
time and then minimize the function.
v1t
A
d
L
v2t
t=
B
At time t the distances moved by A
and B are v1t and v2t respectively.
⇒
Lmin = 5 2 km Illustration 67
The distance between two moving particles P and Q
at any time is a . If vr be their relative velocity and if
u and v be the components of vr along and perpendicular to PQ , then show that their closest distance
av
is
and that the time that elapses before they arrive
vr
au
at their nearest distance is 2 .
vr
Solution
METHOD I
Assuming P to be at rest, particle Q is moving
with velocity vr in the direction shown in figure.
Components of vr along and perpendicular to PQ
are u and v respectively. In the figure
sin a =
u
v
, cos a =
vr
vr
P
Their separation L after time t is given by
L = ( d - v1t )
2
2
α
+ v22 t 2 …(1)
Differentiating w.r.t. time
2L
⇒
( v12 + v22 ) t = v1d ⇒
t=
t=
R
u
vr
dL
= 2 ( d - v1t ) ( -v1 ) + 2v22 t
dt
When the separation L is minimum,
⇒
a
(
v1 d
2
v1 + v22
)
( 10 )( 20 )
( 20 + 20 ) 04_Kinematics 1_Part 2.indd 66
2
2
dL
=0
dt
Q
α
v
(i) The closest distance between the particles is PR .
So,
⎛ v⎞
smin = PR = PQ cos a = ( a ) ⎜ ⎟
⎝ vr ⎠
⇒
smin =
av
vr
11/28/2019 7:50:58 PM
Chapter 4: Kinematics I
(ii) Time after which they arrive at their nearest
­distance is
t=
QR ( PQ ) sin a
=
=
vr
vr
( a ) ⎛⎜ u ⎞⎟
⎝ vr ⎠ au
= 2
vr
vr
METHOD II
In time t , particle P will travel a distance PR = ut
relative to Q along PQ and a distance RS = vt perpendicular to PQ . S is the position of P after time
t relative to Q . The distance x between them after
time t is given by
2
x 2 = QR2 + RS2 = ( a - ut ) + v 2 t 2
x 2 = ( u2 + v 2 ) t 2 - 2 aut + a 2
vsr = vs - vr
⇒ vs = vsr + vr If vr = 0 , then vs = vsr
Conceptual Note(s)
If vr = 0, then v s = v sr
In other words, velocity of man in still water = velocity
of man w.r.t. river.
So, whenever a problem says that velocity of
swimmer in still water is 5 kmhr -1, then we must take
v sr = 5 kmhr -1.
{∵ u2 + v2 = vr2 }
x 2 = vr2 t 2 - 2 aut + a 2 River Problem in One Dimension
Q
α
a
Velocity of river is u and velocity of man in still
water is v .
R
S
CASE-1
Man swimming downstream (along the direction of
river flow) In this case
P
+
Swimmer
vsr = v
This can also be written as
x
2
au ⎞
⎛
= vr2 ⎜ t - 2 ⎟
vr ⎠
⎝
2
+
a2 v2
vr2
…(1)
From equation (1) it is clear that x is minimum at
au
t = 2 and the minimum value of x is
vr
xmin =
av
vr
Relative Motion In River Flow
(One-Dimensional Approach)
If a swimmer ( s ) can swim relative to river ( r ) with
velocity vsr and river is flowing relative to ground
with velocity vr , velocity of swimmer relative to the
ground is vs then
04_Kinematics 1_Part 2.indd 67
4.67
vr = u
Observer at ground sees the swimmer
moving with velocity vs = (v + u)
Velocity of river vr = + u
Velocity of man w.r.t. river vsr = + v
Now vs = vsr + vr = u + v
CASE-2
Man swimming upstream (opposite to the direction
of river flow). In this case
+
Swimmer
vsr = v
vr = u
Observer at ground sees the swimmer
moving with velocity vs = (v – u)
11/28/2019 7:51:03 PM
4.68 JEE Advanced Physics: Mechanics – I
let vsr make an angle q with PQ shown in the figure.
Let us assume the flow of river to be along the positive x-axis.
Since from (1), we get
vs = vsr + vr
Velocity of river vr = -u
Velocity of man w.r.t. river vsr = + v
Now vs = vsr + vr = ( v - u )
Illustration 68
A swimmer capable of swimming with velocity v
relative to water jumps in a flowing river having
velocity u . The man swims a distance d downstream
and returns to the original position. Find out the time
taken in complete motion.
⇒ ( vs )x = ( vsr )x + ( vr )x
⇒ ( vs )y = ( vsr )y + ( vr )y
⇒ ( vs )x = ( -vsr sin q ) + vr
⇒ ( vs )y = vsr cos q + 0
Solution
If t is the time taken by the swimmer to cross the
river, then
⎛ Time of ⎞ ⎛ Time of ⎞
Total time = ⎜ swimming ⎟ + ⎜ swimming ⎟
⎜⎝ downstream ⎟⎠ ⎜⎝ upstream ⎟⎠
t = tdown + tup =
2dv
d
d
+
=
v + u v - u v 2 - u2
CATEGORY 2: RIVER-BOAT PROBLEMS OR
RIVER-SWIMMER PROBLEMS
While solving problems related to river boat or river
swimmer we come across the ­following terms
(a) Absolute velocity of swimmer/boatman vs or
vb (absolute means velocity relative to ground)
(b) Absolute velocity of river current ( vr ) and
(c) Velocity of swimmer/boatman with respect to
the river vsr or vbr
Q
l
y
vr
P
t=
⇒
t=
l
( vs ) y
l
…(4)
vsr cos q
Further if x is the drift (displacement along x axis)
when he reaches the opposite bank, then
x = ( vb )x t
⇒
x = Drift = ( vr - vsr sin q )
x
l
vsr cos q
For t to be MINIMUM, cos q must be MAXIMUM
i.e., cos q = 1
⇒
q = 00 Since, vsr = vs - vr
⇒ vs = vsr + vr …(1)
Observing the situation shown in the diagram below,
we arrive at some standard results and their special
cases.
Consider a river of width l across which a
swimmer wants to swim. Let vsr be the relative
velocity of swimmer with respect to river current and
04_Kinematics 1_Part 2.indd 68
l
…(5)
vsr cos q
Condition for the Swimmer to
Cross the River in the Minimum
Possible Time
Since t =
vsr
θ
⇒ ( vs )x = vr - vsr sin q …(2) ⇒ ( v ) = v cos q …(3)
s y
sr
Q
B
vsr
vs
P
vr
i.e., The relative velocity of swimmer with respect
to river must be perpendicular to the river current.
11/28/2019 7:51:08 PM
Chapter 4: Kinematics I
As a result of this the swimmer will be drifted to the
point B.
So,
⎛ v ⎞
AB = Drift = vr t = ⎜ r ⎟ l …(6)
⎝ vsr ⎠
Condition for Zero Drift
or Condition to Reach the
Opposite Point
So, we conclude that the swimmer reaches Q when
he directs his velocity relative to river at an angle
⎛ v ⎞
q = sin -1 ⎜ r ⎟ , upstream from PQ .
⎝ vsr ⎠
Also we observe that if vr > vsr then the swimmer
would never reach the point Q.
Condition for Minimum Drift
For minimising the drift, we must have
Since we calculated the drift value to be x given by
x = ( vr - vsr sin q )
l
vsr cos q
dx
=0
dq
⇒
d ⎡
l
⎤
=0
( vr - vsr sin q )
⎢
dq ⎣
vsr cos q ⎥⎦
⇒
d ⎡ ⎛ vr
⎞ ⎤
sec q - tan q ⎟ l ⎥ = 0
⎢
⎜
dq ⎣ ⎝ vsr
⎠ ⎦
⇒
vr
( sec q tan q ) - sec2 q = 0
vsr
⇒
vr
( tan q ) - sec q = 0
vsr
⇒
vr sin q
1
=
vsr cos q cos q
For the above condition to be met, we must have
x=0
⇒
vr = vsr sin q ⇒
sin q =
vr
vsr
⎛ v ⎞
⇒ q = sin -1 ⎜ r ⎟ …(7)
⎝ vsr ⎠
A diagramatic representation is given in support of
the above mathematical argument.
vr
B
vsr
Q
B
vs
θ
θ
04_Kinematics 1_Part 2.indd 69
BQ vr
=
PB vsr
⇒
-1 ⎛ vsr ⎞
Total angle =
vr
Let vsr be the velocity of swimmer relative to river
current, vr the river current, then the swimmer
wants to swim in a manner such that he reaches the
point just opposite to the point from where he started.
For such a thing to happen he must have his relative
velocity directed in a direction opposite to the river
current at an angle q to the line PQ (say).
Further since
vsr = vs - vr ⇒ vs = vsr + vr
So, sin q =
If we are asked to calculate the angle with the horizontal, then we get
vs
P
P
⇒ q = sin ⎜
⎝ vr ⎟⎠ Q
vsr
4.69
⎛ v ⎞
q = sin -1 ⎜ r ⎟
⎝ vsr ⎠
π
π
⎛v ⎞
+ q = + sin -1 ⎜ sr ⎟
2
2
⎝ vr ⎠
Also, we conclude that the drift x can be minimised
only if vsr < vr .
Illustration 69
A swimmer can swim at the rate of 5 kmh -1 in still
water. A 1 km wide river flows at the rate of 3 kmh -1 .
The swimmer wishes to swim across the river directly
opposite to the starting point.
(a) Along what direction must the swimmer swim?
(b) What should be his resultant velocity?
(c) How much time will he take to cross the river?
11/28/2019 7:51:13 PM
4.70 JEE Advanced Physics: Mechanics – I
From equations (1) and (2) we get
Solution
(a) The velocity of swimmer with respect to river
vsr = 5 kmh -1 , this is greater than the river flow
velocity, therefore, he can cross the river directly
(along the shortest path). The angle of swim
must be
π
⎛ v ⎞
⎛ v ⎞
q = + sin -1 ⎜ r ⎟ = 90° + sin -1 ⎜ r ⎟
v
2
⎝ sr ⎠
⎝ vsr ⎠
⎛ 3⎞
q = 90° + sin -1 ⎜ ⎟ = 90° + 37°
⎝ 5⎠
2
vs = vsr
- vr2 = 52 - 3 2 = 4 kmh -1
Along the direction perpendicular to the river
flow.
(c) Time taken to cross the
d
=
2
vsr
- vr2
1 km
4 kmh
-1
=
y
⇒
1
h = 15 min
4
Illustration 70
The current velocity of river grows in proportion to
the distance from its bank and reaches the maximum
value v0 in the middle. Near the banks the velocity is
zero. A boat is moving along the river in such a manner that it is always perpendicular to the current. The
speed of the boat in still water is u . Find the distance
through which the boat crossing the river will be carried away by the current if the width of the river is l .
Also determine the trajectory of the boat.
∫
0
⇒
At
q = 127° w.r.t. the river flow or 37° w.r.t. perpendicular in backward direction
(b) Resultant velocity will be
t=
dy
ul
=
dx 2v0 y
⇒
x
ul
ydy =
dx
2v0
∫
0
lv
l
y= , x= 0
2
4u
xnet = 2x =
Condition When the Boatman
Crosses the River Along the
Shortest Route
Here we have to discuss this condition in the light of
two cases.
CASE-1: When the velocity of the swimmer with
respect to river ( vsr ) is greater than river current
( vr ) .
CASE-2: When the velocity of the swimmer with
respect to river ( vsr ) is less than river current ( vr ) .
For the sake of ease we re-assign symbols to vsr and
vr as v and u respectively.
Q
dx ⎛ 2v0 ⎞
vr = vx =
=⎜
⎟ y …(2)
dt ⎝ l ⎠
vbr
y
vr
l
R
L
vsr
Solution
04_Kinematics 1_Part 2.indd 70
lv0
2u P
dy
Given that vbr = vy =
= u …(1)
dt
ulx
y =
v0
2
vs
θ
l = width of
river
β
vr
Let us further assume that vsr (i.e., v ) makes angle q
with the river current vr ( i.e., u ) and vs makes angle
b with the river current. Then,
tan b =
vsr sin q
v sin q
=
…(1)
vr + vsr cos q u + v cos q
Total length of the path is PR = L. Further
y
x
L = PR =
l
= lcosec b
sin b
11/28/2019 7:51:21 PM
Chapter 4: Kinematics I
⇒
L = l 1 + cot 2 b ⇒
⎛ u + v cos q ⎞
L = l 1+ ⎜
⎝ v sin q ⎟⎠
⇒ tan b =
2
⇒
L=
l
u2 sin 2 q + u2 + v 2 cos 2 q + 2uv cos q
v sin q
⇒
L=
l
u2 + v 2 + 2uv cos q
v sin q
v 1 - cos 2 q
u + v cos q
v2
u2
⇒ tan b =
⎛ v⎞
u + v⎜ - ⎟
⎝ u⎠
2
For L to be MINIMUM or L to be MINIMUM
⇒ b = tan -1
⇒
d ⎡ l 2 ⎛ u2 + v 2 + 2uv cos q ⎞ ⎤
⎢ ⎜
⎟⎠ ⎥ = 0
dq ⎣ v 2 ⎝
sin 2 q
⎦
sin 2 q ( -2uv sin q ) -
( u2 + v2 + 2uv cos q ) ( 2 sin q cos q )
⇒
⇒
⇒
⇒
⇒
⇒
sin 4 q
=0
uv sin 2 q + ( u2 + v 2 + 2uv cos q ) cos q = 0 uv ( 1 - cos 2 q ) + ( u2 + v 2 + 2uv cos q ) cos q = 0 uv - uv cos 2 q + ( u2 + v 2 ) cos q + 2uv cos 2 q = 0 uv cos 2 q + ( u2 + v 2 ) cos q + uv = 0 cos q = -
⇒
cos q = -
⇒
cos q = -
( u 2 + v 2 ) ± ( u 2 + v 2 )2 - 4 u 2 v 2
2uv
( u2 + v 2 ) ± ( u2 + v )
2uv
( u2 + v 2 ) ± ( u2 + v 2 )
2uv
Either
cos q =
2 2
2
2
-u - v + u - v
2uv
⇒ cos q = -
2
v
…(2)
u
Equation (2) holds good
only when v < u
Further put (2) and (3)
one by one in (1), then
For v < u we have
v
cos q = u
04_Kinematics 1_Part 2.indd 71
u -v
v
2
u -v
⎞
2 ⎟
⎠
⇒ b=
π
2
So, we conclude that the direction of shortest route
followed by the swimmer is at right angles to the
river ­current when velocity of swimmer with respect
to river current (v) is greater then river current ( vr )
or u and for v < u (the opposite case) it is in a direcv
⎛
⎞
tion b = tan -1
⎜⎝
2
2 ⎟
u -v ⎠
CATEGORY 3: AEROPLANE-WIND
PROBLEMS
These problems proceed the same way as we have
done for river-swimmer problems or river-boat
­problems. Here vsr is just replaced by vaw (velocity
of aircraft with respect to wind), where vaw = va - vw .
Illustration 71
-u2 - v 2 - u2 + v 2
2uv
A plane moves in windy weather due east while the
pilot points the plane somewhat south of east. The
wind is blowing at 50 kmhr -1 directed 30° east of
north, while the plane moves at 200 kmhr -1 relative
to the wind. What is the velocity of the plane relative
to the ground and what is the direction in which the
pilot points the plane?
cos q =
⇒ tan b → ∞
2
or
2
v 1-
v
2
⎛
⎜⎝
v 1 - cos 2 q
u + v cos q
u2
v2
⇒ tan b =
⎛ u⎞
u + v⎜ - ⎟
⎝ v⎠
v 1-
⇒ tan b =
d ( 2)
L =0
dq
⇒ tan b =
4.71
u
…(3)
v
Equation (3) holds
good only when v < u
⇒ cos q = -
For u < v we have
u
cos q = v
Solution
Three vectors vPG , vPW and vWG are involved. We
have to find the magnitude of vPG and the direction
of vPW (i.e., q ). Let us first relate the three vectors
using the concept of relative velocity. We can write
⎛ Velocity of ⎞ ⎛ Velocity of ⎞ ⎛ Velocity of ⎞
⎜
⎟ -⎜
⎟
plane ⎟ = ⎜
plane
wind
⎜⎝ w.r.t. wind ⎟⎠ ⎜⎝ w.r.t. ground ⎟⎠ ⎜⎝ w.r.t. ground ⎟⎠
11/28/2019 7:51:34 PM
4.72 JEE Advanced Physics: Mechanics – I
θ
vPW
⇒
⇒
vPG
y
vWG
N
x
Here we have two unknown quantities ( vPG and q ) ,
so we should not relate the magnitudes of these vectors using triangle law. In such situations we resolve
the vectors into components on the coordinate system and then solve the equations for both axes
( x and y ) .
For the y components
vPG , y = vPW , y + vWG , y
i.e., 0 = - ( 200 ) sin q + ( 50 cos 30° )
Solving, we get
3
; q = sin -1 ( 0.216 ) = 12°
8
For the x components
vPG , x = vPW , x + vWG , x
45° α
vPG = 221 kmh -1 METHOD I
Given that vw = 200 2 kmhr -1 , vaw = 400 kmhr -1
and va should be along AB or in north-east direc
tion. Thus, the direction of vaw should be such as the
resultant of vw and vaw is along AB or in north-east
direction.
04_Kinematics 1_Part 2.indd 72
E
AC
BC
=
sin 45° sin a
⎛ 200 2 ⎞ 1
1
⎛ BC ⎞
sin a = ⎜
sin 45° = ⎜
=
⎝ AC ⎟⎠
⎝ 400 ⎟⎠ 2 2
⇒ a = 30° ⇒
Therefore, the pilot should steer in a direction at an
angle of ( 45° + a ) or 75° from north towards east.
va
400
=
Further,
sin ( 180° - 45° - 30° ) sin 45°
⇒
sin 105°
va =
× ( 400 ) kmh -1
sin 45°
⇒
⎛ cos 15° ⎞ (
va = ⎜
400 ) kmh -1
⎝ sin 45° ⎟⎠
0
.
9659
⎛
⎞(
va = ⎜
400 ) kmh -1 = 546.47 kmh -1
⎝ 0.707 ⎟⎠
So, the time of journey from A to B is
AB
1000
t= =
hr
va
546.47
An aircraft flies at 400 kmhr -1 in still air. A wind of
Solution
vaw = 400 kmhr –1
Let vaw makes an angle a with AB as shown in
­figure. Applying sine law in triangle ABC , we get
Illustration 72
200 2 kmhr -1 is blowing from the south. The pilot
wishes to travel from A to a point B north east of
A. Find the direction he must steer and time of his
journey if AB = 1000 km .
C
A
vPG = ( 200 cos q ) + ( 50 sin 30° )
⇒
45° v = 200 2 kmhr –1
w
va
vPW = vPG - vWG vPG = vPW + vWG sin q =
B
⇒
t = 1.83 hr METHOD II
Suppose a vector C is a vector sum of two vectors
the
direction
of
is given to us. Let
A and B and
C
the vector C be directed along a line PQ . Then
the line PQ , i.e., the
A + B ( = C ) should be along
sum of components of A and B perpendicular to
line PQ must be zero. Similarly, if C = A - B and C
is directed along the line PQ , then the sum of components of A and -B perpendicular to line PQ must
be zero.
11/28/2019 7:51:47 PM
Chapter 4: Kinematics I
For example, if va has to be along AB and we know
that va = vaw + vw . Therefore, sum of components of
vaw and vw perpendicular to line AB (shown as dotted) should be zero.
N
va = v + u
⇒
vw
45°
⇒
E
tRS =
⇒
tSP =
⇒
t=
⇒
t=
Hence, the time of journey from A to B is given by
AB
1000
t= =
= 1.83 hr
va
546.47
Find the time an aeroplane having velocity v , takes
to fly around a square with side l if the wind is blowing at a velocity u along one side of the square.
Solution
Velocity of aeroplane while flying from P to Q is
04_Kinematics 1_Part 2.indd 73
l
v-u va = v – u
u
v
l
va = v 2 – u2
u
v 2 - u2 t = tPQ + tQR + tRS + tSP
l
l
l
l
+
+
+
2
2
2
v-u
v+u
v -u
v - u2 2a
2
v - u2
( v + v2 - u2 )
CATEGORY 4: RAIN-MAN PROBLEMS
Illustration 73
R
Q
v 2 - u2
va = v 2 - u 2
3
+ 200
va = ( 400 )
2
va = 346.47 + 200 = 546.47 kmhr -1
P
va = v 2 – u2
Velocity of aeroplane while flying from S to P is
Total time
vwind = u
v
v
⎛ 1 ⎞
va = ( 400 ) cos 30° + ( 200 2 ) ⎜
⎝ 2 ⎟⎠
S
l
Velocity of aeroplane while ­flying from R to S is
⇒
⇒ a = 30° Now, va = vaw cos a + vw cos 45°
⇒
u
va = v - u
So, vaw sin a = vw sin 45°
vw
⇒ sin a = sin 45°
vaw
⎛ 200 2 ⎞ ⎛ 1 ⎞ 1
⇒ sin a = ⎜
=
⎝ 400 ⎟⎠ ⎜⎝ 2 ⎟⎠ 2
⇒
tQR =
vaw
α
l
v+u va = v 2 - u 2
A
⇒
tPQ =
Velocity of aeroplane while flying from Q to R is
B
va
4.73
va = v + u
In such problems we come across the following
­terminology according to which
(a) vr is the absolute velocity of rain.
(b) vm is the absolute velocity of man/cyclist/
motorist/observer.
(c) vw is the absolute velocity of wind (if it is
­blowing) and
(d) vrm is the velocity of rain with respect to man or
the velocity of rain which appears to the man.
For dealing with such like problems, we have two
options
11/28/2019 7:51:56 PM
4.74 JEE Advanced Physics: Mechanics – I
Option A: When no wind is blowing.
vrm = vr - vm or vrm = vr + ( -vm )
Illustration 74
A standing man, observes rain falling with velocity of
20 ms -1 at an angle of 30° with the vertical.
(a) Find the velocity with which the man should
move so that rain appears to fall vertically to
him.
(b) Now if he further increases his speed, rain again
appears to fall at 30° with the vertical. Find his
new velocity.
Solution
(a) vm = -viˆ (let)
v = -10iˆ - 10 3 ˆj
r
v = - ( 10 - v ) iˆ - 10 3 ˆj
rm
04_Kinematics 1_Part 2.indd 74
0m
s –1
30
°
=2
vr
⇒ - ( 10 - v ) = 0 (for vertical fall, horizontal
component must be zero)
-1
⇒ v = 10 ms
(b) vr = -10iˆ - 10 3 ˆj
v
v ˆ
m = - x i
vrm = - ( 10 - vx ) iˆ - 10 3 ˆj
°
30
30
°
s –1
10
0m
So, while dealing with the problems involving rain,
man and the blowing wind we first calculate the resultant of rain and wind i.e., net velocity of rain under the
inference of wind vr + vw = vnet rain . The man sees this
rain falling with a velocity vrm = vnet rain - vm
⇒ vrm = ( vr + vw ) - vm
⇒ vrm = ( vr + vw ) + ( -vm )
Again, we reverse the direction of velocity of man
and then find the resultant of vnet rain and -vm to get
vrm with magnitude and direction.
So, to deal with problems involving rain, man
and wind we just reverse the direction of vm i.e.,
make it -vm and then find the resultant of vr , vw
and -vm i.e., vr + vw + ( -vm ) .
vm
=2
Option B: When wind is blowing.
vrm = ( vr + vw ) - vm or vrm = vr + vw + ( -vm )
vrm
10 3 ms–1
vr
So, while dealing with the problems in which rain and
man are there but no wind exists, then to calculate the
direction of vrm , we simply reverse the direction of
man’s velocity ( -vm ) and then find the resultant of
-vm and vr i.e., -vr + ( -vm ) to get the direction and
magnitude of vrm . It’s the Best Trick! Try following
and see the results.
10
vrm
10 3
60°
60°
vm
Angle with the vertical = 30°
10 - vx
⇒
tan 30° =
⇒
vx = 20 ms -1 -10 3 Illustration 75
To a man walking at the rate of 3 kmh -1 the rain
appears to fall vertically. When he increases his speed
to 6 kmh -1 it appears to meet him at an angle of 45°
will vertical. Find the speed of rain.
Solution
Let î and ĵ be the unit vectors along the horizontal
and vertical directions respectively.
Let velocity of rain be vr = aiˆ + bjˆ …(1)
Then speed of rain will be
vr = a 2 + b 2 …(2)
11/28/2019 7:52:06 PM
Chapter 4: Kinematics I
In the first case vm = velocity of man = 3î
⇒ vrm = vr - vm = ( a - 3 ) iˆ + bjˆ
It seems to be in vertical direction. Hence, the horizontal component must be zero.
⇒
⇒
a-3 = 0
a=3
4.75
This seems to be at 45° with the vertical
⇒
tan ( 45° ) =
⇒
b =3
vy
vx
=
b
3
Therefore, from equation (2) speed of rain is
2
2
vr = ( 3 ) + ( 3 ) = 3 2 kmh -1
In the second case vm = 6iˆ
⇒ vrm = ( a - 6 ) iˆ + bjˆ = -3iˆ + bjˆ
Test Your Concepts-VII
Based on Relative velocity
(Solutions on page H.86)
1. A river 400 m wide is flowing at a rate of 2 ms -1. A
boat is sailing at a velocity of 10 ms -1 with respect
to the water, in a direction perpendicular to the
river.
(a) Find the time taken by the boat to reach the
opposite bank.
(b) How far from the point directly opposite to
the starting point does the boat reach the
opposite bank?
2. Snow is falling vertically at a constant speed of
8 ms -1. At what angle from the vertical do the
snow flakes appear to be falling as viewed by the
driver of a car travelling on a straight, level road
with a speed of 50 kmh-1 ?
3. Show that the direction of shortest route is at
right angles to the river when the velocity of
boat with respect to water v is greater than that
of the river velocity u and in opposite case it is
v
⎛
⎞
tan-1
.
⎜⎝ 2
2 ⎟
u -v ⎠
4. A ship A is travelling due east at 10 kmhr -1 and at
9 am is 30 km south-west of another ship B. If B
travels at 15 kmhr -1 so as to intercept A calculate
the
(a) direction in which B must travel.
(b) time it takes when the interception takes
place.
5. A motorboat going downstream overcame a raft at
point A. One hour later it turned back and met the
raft again at a distance 6 km from point A. Find the
river velocity.
04_Kinematics 1_Part 2.indd 75
6. An elevator car whose floor to ceiling distance
is equal to 2.7 m starts ascending with constant
acceleration 1.2 ms -2 . 2 second after the start, a
bolt begins falling from the ceiling of the car. Find
the
(a) time after which bolt hits the floor of the
elevator.
(b) net displacement and distance travelled by the
bolt, with respect to earth.
( Take g = 9.8 ms-2 )
7. A particle is moving in a circle of radius r centred
at O with constant speed v. Calculate the change in
velocity in moving from A to B when ∠AOB = 40°.
8. Two ships A and B are 10 km apart on a line
running south to north. Ship A farther north
is streaming west at 20 kmhr -1 and ship B is
streaming north at 20 kmhr -1. What is their distance of closest approach and how long do they
take to reach it?
9. Car A has an acceleration of 2 ms -2 due east and
car B, 4 ms -2 due north. What is the acceleration
of car B with respect to car A?
10. A bullet train A starts from rest at t = 0 and travels along a straight track with a constant acceleration of 6 ms -2 until it reaches a speed of 80 ms -1.
Afterwards it maintains this speed. Also, when
t = 0, another bullet train B located 6000 m down
on a parallel track is travelling towards A at a constant speed of 60 ms -1. Determine the distance
travelled by train A when they pass each other.
11/28/2019 7:52:11 PM
4.76 JEE Advanced Physics: Mechanics – I
11. A body is thrown up in a lift with a velocity u relative to the lift. If the time of flight is found to be t
then find the acceleration of the lift.
12. A man wants to reach point B on the opposite bank
of a river flowing at a speed as shown in the figure.
What minimum speed relative to water should the
man have so that the can reach point B? In which
direction should he swim?
B
u
45°
A
13. A launch plies between two points A and B on
the opposite banks of a river always following the line AB. The distance S between points
A and B is 1200 m. The velocity of the river current v = 1.9 ms -1 is constant over the entire width
of the river. The line AB makes an angle a = 60°
with the direction of the current. With what velocity u and at what angle b to the line AB should the
launch move to cover the distance AB and back in
a time t = 5 min? The angle b remains the same
during the passage from A to B and from B to A.
B
u
β
α
v
A
14. A man wishes to cross a river of width 120 m by
a motorboat. His rowing speed in still water is
3 ms -1 and his maximum walking speed is 1 ms -1.
The river flows with velocity of 4 ms -1.
(a)Find the path which he should take to get to
the point directly opposite to his starting point
in the shortest time.
(b)Also, find the time which he takes to reach his
destination.
15. A ball is thrown vertically upward from the 12 m
level in an elevator shaft with an initial velocity of
18 ms -1. At the same instant an open platform
04_Kinematics 1_Part 2.indd 76
type elevator passes the 5 m level, moving upward
with a constant velocity of 2 ms -1. Find
(a) when and where the ball will hit the elevator
(b)the relative velocity of the ball with respect to
the elevator when the ball hits the elevator.
16. A swimmer heads directly across a river, swimming at 1.6 ms -1 relative to still water. He arrives
at a point 40 m downstream from the point directly
across the river, which is 80 m wide.
(a) What is the speed of the river current?
(b)What is the swimmer’s speed relative to the
shore?
(c)In what direction should the swimmer head so
as to arrive at the point directly opposite to the
starting point?
17. A cyclist, riding at a speed V, overtakes a pedestrian
who can move at a speed not greater than v, the
two travelling along parallel tracks at a distance d
apart. Show that if the cyclist rings his bell when
V
at a distance less that d, he may safely maintain
v
his speed and keep to his course regardless of the
behaviour of the pedestrian.
18. A swimmer wishes to cross a 500 m wide river flowing at 5 kmh-1. His speed with respect to water is
3 kmh-1.
(a)If he heads in a direction making an angle q
with the flow, find the time he takes to cross
the river.
(b)Find the shortest possible time to cross the
river.
(c)Assuming that the swimmer has to reach the
other shore at the point directly opposite to
his starting point. If he reaches the other shore
somewhere else, he has to walk down to this
point. Find the minimum distance that he has
to walk.
19. A pipe which can be rotated in a vertical plane
is mounted on a cart. The cart moves uniformly
along a horizontal path with a velocity v1 = 2 ms -1.
At what angle a to the horizontal should the pipe
be placed so that drops of rain falling vertically with
a velocity v2 = 6 ms -1 move parallel to the walls of
the pipe without touching them.
11/28/2019 7:52:13 PM
Chapter 4: Kinematics I
α
20. Two mirrors, mounted vertically, are made to move
towards each other with a speed v each. A particle
that can bounce back between the two mirrors
04_Kinematics 1_Part 2.indd 77
4.77
starts from one mirror when the mirrors are d
apart. On reaching the second mirror, it bounces
back and so on. If the particle keeps on travelling at
a constant speed of 3 v, how many trips can it make
before the mirrors run into each other? What total
distance does it cover?
11/28/2019 7:52:13 PM
4.78 JEE Advanced Physics: Mechanics – I
Solved Problems
Problem 1
A particle is projected vertically with a speed of
32.9 ms–1 from a point O . It passes successively
through two horizontal thin nets at height 8 m and
16 m respectively, above O . Find the greatest height
above O that the particle attains net and its speed on
reaching back to O . Assume that each thin net has a
mesh which is broad enough to allow the ball to cross
the net but witch half the velocity with which the ball
enters.
∴
Total height above O is 16 + 2.4 = 18.4 m
In the return journey, the velocity of the particle at the
time of reaching net B is given by
2
v32 = ( 0 ) + 2 gh
⇒
v3 = 2 gh = 2 × 9.8 × 2.4
Velocity after crossing net B is
Velocity at net A is
Solution
Take upward direction as positive. In the upward
journey, let v1 be the velocity of the particle just
before it enters the net A . Then
v12 = u12 − 2 gh = (39.2)2 − 2 × 9.8 × 8
8m
O
Solving it, we get v1 = 37.15 ms −1
v
Velocity of the particle after crossing the net is 1 .
2
So
v1
= 18.575 ms −1
2
Let v2 be the velocity of the particle just before it
enters the net B . Then
2
v22 = ( 18.575 ) − 2 × 9.8 × 8
Solving it, we get v2 = 13.72 ms −1
Velocity of the particle after it crosses the net B is
v2 =
13.72
= 6.86 ms −1
2
Let h be the height reached above the net B where
velocity of particle becomes zero. Then
2
2
( 0 ) = ( 6.86 ) − 2 × 9.8 × h
⇒ h = 2.4 m
04_Kinematics 1_Part 3.indd 78
9.8 × 17.2
1
9.8 × 17.2
2
Let the velocity at O be v4 . Then
( 9.8 × 17.2 )
4
⇒ v4 = 14.1 ms −1
16 m
⎛ 2 × 9.8 × 2.4 ⎞
⎜⎝
⎟⎠ + 2 × 9.8 × 8
4
Velocity after crossing the net A is
v42 =
Net B
Net A
u1
( vA )down =
1
2 × 9.8 × 2.4
2
+ 2 × 9.8 × 8
Problem 2
Find the trajectory of the particle of mass m acted
upon by a force
kˆ F = cos tiˆ + sin tjˆ . At t = 0 , the position
and velocity of particle are x iˆ and v ˆj respectively.
0
0
Solution
d2 r
Since,
kˆ F = m 2 = cos tiˆ + sin tjˆ {Given}
dt
Integrating both sides w.r.t. t , we have
⎛ d2 r ⎞
kˆ ⎜⎜ m 2 ⎟⎟ dt = (cos tiˆ + sin tjˆ )dt
⎝ dt ⎠
dr
ˆ
= sin tiˆ + (− cos t) ˆj + c1 …(1)
⇒k m
dt
At time t = 0 , we have
…(2)
∫
∫
mv0 ˆj = − ˆj + c1 {putting t = 0 in equation (1)}
⇒
c1 = ( mv0 + 1 ) ˆj
Substituting value of c1 in equation (1) we get
dr
kˆ m
= sin tiˆ + (− cos t) ˆj + ( mv0 + 1 ) ˆj
dt
11/28/2019 7:08:32 PM
Chapter 4: Kinematics I
⇒ kˆ m
dr
= sin tiˆ + ( mv0 + 1 − cos t ) ˆj
dt
u cos θ
c2 = ( mxo + 1 ) iˆ
Substituting value of c2 in equation (3), we get
kˆ mr = ( mxo + 1 − cos t ) iˆ + ( mvo t + t − sin t ) ˆj
1
⇒ kˆ r = ⎡⎣ ( mxo + 1 − cos t ) iˆ + ( mvo t + t − sin t ) ˆj ⎤⎦
m
…(4)
Comparing equation (4) with r = xiˆ + yjˆ
\
θ
∫
Again at time t = 0, r = xo iˆ, thus equation (3) gives us
⎛ 1 − cos t ⎞
⎛ t − sin t ⎞
x = xo + ⎜
⎟ and y = vo t + ⎜⎝
⎟
⎝
m ⎠
m ⎠
A sitting cat in a field suddenly sees a standing dog.
To save its life the cat runs away in a straight horizon
tal line with constant velocity u . Without any time
lag the dog starts with velocity v constant in magnitude and always directed towards the cat to catch it.
Initially, v and u are perpendicular and separation
between the cat and the dog is d . Find the time after
which the dog catches the cat.
B
u
makes an angle q with u . Then resolving u into two
components
(a) u cos θ , parallel to v , and
(b) u sin θ , perpendicular to v .
Therefore, relative velocity of approach of A towards
B is
dx
(v − u cos θ ) = −
dt
(Negative sign indicates that distance between particles decreases with time)
⇒ − dx = (v − u cos θ )dt …(1)
Integrating both sides of equation (1),
t
⇒
∫
∫
− dx = (v − u cos θ ) dt
0
d
t
⇒
d=
∫
t
∫
v dt − u cos θ dt …(2)
0
0
Since u < v , Therefore, u cos θ < v and hence, they
meet.
Also, the horizontal distance travelled by both of
them is the same, when they meet.
∫ v dt
x
t
v
A
⇒
ut =
Solution
Initially the particles are at A and B, a distance d apart
and their velocities are perpendicular to each other.
The velocity u is constant in magnitude and
direction while the velocity v changes continuously
in direction and is always aimed at B. Let, at any
instant t , v
∫ v cosθ dt
0
At t = 0
04_Kinematics 1_Part 3.indd 79
u sin θ
At any time t (say)
Therefore, Δx =
d
v cos θ = vx
u
A
0
Problem 3
θ
v
Integrating again both sides w.r.t. t , we have
⎛ dr ⎞
kˆ ⎜ m ⎟ dt = (sin tiˆ + ( mv0 + 1 − cos t ) ˆj ) dt
⎝ dt ⎠
⇒ kˆ mr = ( − cos t ) iˆ + ( mvo t + t − sin t ) ˆj + c2 …(3)
∫
4.79
t
⇒
∫
t
∫
d = v dt − u cos θ dt
0
0
⇒
⎛ ut ⎞
d = vt − u ⎜ ⎟
⎝ v⎠
⇒
t=
vd
v 2 − u2
11/28/2019 7:08:39 PM
4.80 JEE Advanced Physics: Mechanics – I
Problem 4
A hinged construction consists of three rhombus
with ratio of sides 3 : 2 : 1 (as shown in figure). Vertex
A3 moves in a horizontal direction with velocity v.
Determine velocities of vertices A1 , A2 and B2 at
instant when angle of construction is 45° .
B1
⇒
θ
A1
A2
C2
A3
v
B2 x
Solution
Let A0 be the Origin of the whole system shown in
the diagram. Let
A0 B1 = B1 A1 = 3 k
A1B2 = B2 A2 = 2k
A2 B3 = B3 A3 = k
Hence, A0 A1 = 3 K cos θ + 3 K cos θ = 6 K cos θ = x A1
A0 A2 = A0 A1 + A1 A2 = 6 K cos θ + 4 K cos θ
⇒
x A2 = 10 K cos θ
A0 A3 = A0 A2 + A2 A3 = 10 K cos θ + 2K cos θ
⇒
x A3 = 12K cos θ
Similarly,
yB2 = 2K sin θ
According to Problem,
⇒
−12K sin θ
⇒
− K sin θ
dx A3
dt
=v
dθ
=v
dt
v
dθ
=
dt 12
Therefore,
vA2 =
04_Kinematics 1_Part 3.indd 80
B2 y
( ) = (v ) + (v )
2
2
B2 x
2
B2 y
2
⇒
( v ) = 64 ⎛⎜⎝ K sin θ ddtθ ⎞⎟⎠ + 4 cos θ ⎛⎜⎝ K ddtθ ⎞⎟⎠
⇒
( v ) = 64 ⎛⎜⎝ 12v ⎞⎟⎠ + 4 cos θ ⎛⎜⎝ K ddtθ ⎞⎟⎠
⇒
( )
2
⇒
( )
2
⇒
vB2 =
2
2
B2
2
2
2
B2
vB2
vB2
2
2
⎛ ⎛ v ⎞2 1 ⎞
⎛ v ⎞
= 64 ⎜ ⎟ + 4 cos 2 θ ⎜ ⎜ ⎟
2 ⎟
⎝ 12 ⎠
⎝ ⎝ 12 ⎠ sin θ ⎠
⎛ v ⎞
=⎜ ⎟
⎝ 12 ⎠
2
( 64 + 4 cot θ )
2
v
64 + 4 cot 2 θ
12
Now when θ is 45° , then
⇒ vB2 =
v
17
68 =
v
12
6
Problem 5
xB2 = 6 K cos θ + 2K cos θ = 8 K cos θ ,
and
dyB2
( v ) = dt = 2K cos θ ⎛⎜⎝ ddtθ ⎞⎟⎠
Now, vB2
C3
C1
dxB2
( v ) = dt = −8K sin θ ⎛⎜⎝ ddtθ ⎞⎟⎠
B3
A0
⎛ dθ ⎞
⎛ v ⎞ v
= −6 K sin θ ⎜
⎟⎠ = 6 ⎜⎝ ⎟⎠ =
⎝
dt
dt
12
2
For B2 :
⇒
B2
dx A1
vA1 =
dx A2
⎛ dθ ⎞
⎛ v ⎞ 5
= −10 K sin θ ⎜
= 10 ⎜ ⎟ = v
⎟
⎝ dt ⎠
⎝ 12 ⎠ 6
dt
At the initial moment three points A, B, and C are on
a horizontal straight line at equal distance from one
another. Point A begins to move vertically upwards
with a constant velocity v and point C vertically
downward without any initial velocity at a constant
acceleration a . How should the point B move vertically for all the three points to be constantly on one
straight line? The points begin to move simultaneously.
Solution
Initially, at t = 0 , all the particles are equidistant from
each other.
A
B
l
C
l
11/28/2019 7:08:46 PM
Chapter 4: Kinematics I
Let at time t the particle A goes to A ′ , B goes to B ′
and C goes to C ′
AA ′ = vt
BB ′ = uB t +
1 2
aB t
2
1 2
at
2
CC ′ =
straight lines. The boat A along the river and the boat
B across the river. Having moved off an equal distance from the buoy the boats returned. Find the ratio
τ
of the times of motion of boats A if the velocity of
τB
each boat w.r.t. water is h = 1.2 times greater than the
stream velocity.
Solution
In similar triangles AOA′ , BOB′ and COC′ we have
1
1 2
uB t + aB t 2
at
vt
2
=
= 2
…(1)
l−x
x
l+x
⇒
2v
at
=
l−x l+x
⇒
⎛ at − 2v ⎞
x=⎜
l …(2)
⎝ at + 2v ⎟⎠
Let, d is the distance moved by boats, A and B, away
from boys in direction of motion (or perpendicular
to) stream velocity when they go from point 1 to 2.
v is the velocity of stream
Therefore, ηv is the velocity of the boat.
For Boat A:
⎛ Total time ⎞ ⎛ Time taken ⎞ ⎛ Time taken ⎞
⎜ taken by ⎟ = ⎜ downstream ⎟ + ⎜ upstream ⎟
⎟
⎟ ⎜
⎜
⎟ ⎜
⎝ Boat A ⎠ ⎝ by Boat A ⎠ ⎝ by Boat A ⎠
Put (2) in (1) we get
uB t +
1 2 ⎛ −v ⎞
1⎛ a⎞
aB t = ⎜
t + ⎜ ⎟ t2
⎝ 2 ⎟⎠
2
2⎝ 2⎠
i.e., uB = −
4.81
v
2′
v
a
and aB =
2
2
2
η v cos θ
ηv
A′
θ
1
B
O
A
C
l–x
B′
⇒
tA = tdown + tup
⇒
tA =
⇒
tA =
d
d
+
ηv + v ηv − v
x
2dη
v(η2 − 1)
For Boat B: Time taken by Boat B in going from 1 to 2
l+x
Hence, for all the particles to be on one straight line
v
the particle at B must move with an initial velocity
2
a
in the upward direction and an acceleration
in the
2
downward direction.
Problem 6
Two boats, A and B, move away from a buoy anchored
at middle of the river along mutually perpendicular
04_Kinematics 1_Part 3.indd 81
t1→ 2 =
C′
and sinq =
⇒
d
…(1)
ηv cos θ
v 1
=
ηv η
cos θ = 1 −
1
η2
Substituting respective values in (1),
⇒
t1→ 2 =
d
v η2 − 1
11/28/2019 7:08:51 PM
4.82 JEE Advanced Physics: Mechanics – I
2
θ
ηv
1′
η v cos θ
1
v
Since the particle is going from x = 1 m to
x = 0.5 m, hence, its velocity is directed along
the negative x-axis.
⇒kˆ v = (−1)iˆ
(b) v 2 =
Time taken by boat B to return from 2 to 1
⇒
⇒ −v =
d
t2 →1 =
v η2 − 1
⇒ −
2d
⇒
tB = t1→ 2 + t2→1 =
⇒
tA
η
=
= 1.8
tB
η2 − 1
v η2 − 1
k ⎡1
⎤
− 1⎥
m ⎢⎣ x
⎦
t
1
4
0
1
x
∫ dt = −∫ 1 − x dx
Solving, by substituting x = sin 2 θ , we get
A particle of mass 10 −2 kg is moving along the posik
tive x-axis under the influence of force F(x ) = − 2 ,
2x
k = 10 −2 Nm 2 . At t = 0 , it is at x = 1 m and its velocity is v = 0
(a) Find its velocity when it reaches x = 0.5 m .
(b) Time at which it reaches x = 0.25 m .
Solution
(a) F(x ) = −
k
2x 2
{Given}
⎛π
3⎞
t=⎜ +
⎟ s
⎝3
4 ⎠
Problem 8
From point A located on a highway one has to get
by car as soon as possible to point B located in the
field at a distance l from highway. It is known that
car moves in the field η times slower than on highway. At what distance from D one must turn off the
highway?
A
k
dv
⇒ mv
=− 2
dx
2x
⇒ mvdv = −
k
2x 2
v
∫
⇒ m vdv = −
0
k
2
⇒ v = +1
⇒ v = ±1 ms −1
x
D
dx …(1)
B
0.5
∫
x −2 dx
1.0
Solution
Total Time = tA→C + tC →B
⇒
t=
⇒
t=
1
k⎡ 1
⎤
mv 2 = ⎢
− 1⎥
2
2 ⎣ 0.5
⎦
2
C
Highway l
Integrating (1) within suitable limits
04_Kinematics 1_Part 3.indd 82
1− x
dx
…(2)
=
dt
x
Integrating relation (2) within suitable limits
⇒
Problem 7
⇒
1− x
{directed towards negative axis}
x
AC CB
+
v
v
η
x2 + l2
AD − x
+
v
v
η
11/28/2019 7:08:57 PM
Chapter 4: Kinematics I
The car has to reach B as soon as possible
⇒
(b) Let R be the radius of the parabolic equation,
3
dt
=0
dx
∴
⎡ ⎛ dy ⎞ 2 ⎤ 2
⎢1+ ⎜
⎟ ⎥
⎢⎣ ⎝ dx ⎠ ⎥⎦
⇒ R=
⎛ d2 y ⎞
⎜ 2⎟
⎝ dx ⎠
1 1
2x
t=− + ×
v v 2 x2 + l2
η
⇒
η2 x 2
=1
x2 + l2
⇒
x=
3
⎡ ⎛ k x ⎞2 ⎤2
⎢1+ ⎜ 2 ⎟ ⎥
⎢ ⎝ k1 ⎠ ⎥⎦
⇒ R= ⎣
⎛ k2 ⎞
⎜⎝ k ⎟⎠
1
l
2
η −1
Problem 9
A particle moves in x-y plane with velocity
v = k1iˆ + k 2 xjˆ , where î and ĵ are unit vectors along
x and y axes, k1 and k 2 are constants. Initially the
particle is at origin (0, 0). Find
(a) equation of particles trajectory y ( x ) ,
(b) curvature radius of trajectory of particle as a
function of x .
Solution
(a) v = k1iˆ + k 2 xjˆ {Given}
vx = k1iˆ
vy = k 2 xjˆ
dx
= k1
dt
dy
= k 2 x = k1 k 2 t
dt
x
∫
t
dx =
0
∫
k1 dt
0
x = k1t …(1)
y
∫
t
∫
dy = k1 k 2 t dt
0
y=
k ⎡ ⎛ k x⎞
⇒ R = 1 ⎢1+ ⎜ 2 ⎟
k 2 ⎢ ⎝ k1 ⎠
⎣
3
2 ⎤2
⎥
⎥⎦
Problem 10
A helicopter takes off along the vertical with an
acceleration a = 3 ms −2 and zero initial velocity. In
a certain time t1 , the pilot switches off the engine.
At the point of take off, the sound dies away in time
t2 = 30 s . Determine the velocity v of the helicopter at the moment when the engine is switched off;
assuming that the velocity of sound, c = 320 ms −1 .
Solution
0
k1 k 2 2
t …(2)
2
Eliminating t from the above set of two equations, we have
⎛ k ⎞
y = ⎜ 2 ⎟ x 2 …(3)
⎝ 2k1 ⎠
Equation (1) sounds much more like the equation
of the parabola
04_Kinematics 1_Part 3.indd 83
4.83
Since, u = ms–1, a = 3 ms–2, c = 320 ms–1 and t2 = 30 s
{Given}
Velocity of helicopter at time t1 is
v = 0 + 3t1
⇒
v = 3t1
Distance traversed by helicopter in time t1 is
s = 0+
⇒
s=
1
× 3 × t12
2
3 2
t1
2
s
Now, t2 = t1 + {Given}
c
⇒
30 = t1 +
3 t12
2 320
11/28/2019 7:09:04 PM
4.84 JEE Advanced Physics: Mechanics – I
⇒
⇒
3t12 + 640 t1 − 19200 = 0
t1 =
−640 ±
( 640 )2 + 4 × 3 × 19200
6
−640 ± 800
6
As time cannot be negative,
−640 + 800 160 80
⇒ t1 =
=
=
s
6
6
3
⇒
t1 =
⇒
v = 3t1 = 3 ×
⇒
v = 80 ms −1
80
3
Also, from equation of motion v 2 − u2 = 2 as ,
⇒
v12 = 0 + 2 × 10 × h1
v12
( 9.5 )
=
= 4.51 m
2 × 10 2 × 10
⇒ Path above top of Window = 4.51 m
For the path below the bottom of window, the ball
reappears at the bottom of window to second after
passing the bottom on its way down. Hence, it takes
a time of 1 s to fall from the bottom of window to the
ground and rebounds to the same height in a further
time of 1 s.
2
⇒
h1 =
⇒
h2 = v2 t +
⇒
h2 = 10.5 × 1 +
⇒
h2 = 10.5 + 5
Problem 11
A steel ball is dropped from a roof of a building. An
observer standing in front of a window 1 m high
notes that the ball takes 0.1 sec to fall from the top
to the bottom of the window. The ball continues to
fall and makes a complete elastic collision with the
ground, so that it rebounds with the same velocity
with which it strikes the ground. The ball reappears
at the bottom of the window 2 second after passing
the bottom of the window on the way down, find the
height of the building (Take g = 10 ms–2)
Solution
Let v1 and v2 be the velocities of a ball at the top and
bottom of the windows respectively. Let h1 be the
height of the building above the top of the window
and h2 that below the bottom of window, then
Height of Building = h1 + Height of Window + h2
⇒
Height of Building = ( h1 + 1 + h2 ) m
For the path between the top and bottom of window,
We have h = 1 m , t = 0.1 s
a = g = 10 ms −2
h = v1t +
1 2
gt
2
⇒ 1 = v1 × 0.1 +
⇒
1
2
× ( 10 ) × ( 0.1 )
2
v1 = 9.5 ms -1
Also from equation v = u + at
⇒
v2 = v1 + 10 × 0.1 = 9.5 + 1 = 10.5 ms −1
04_Kinematics 1_Part 3.indd 84
1 2
gt
2
1
× 10 × 12
2
⇒ Path below the bottom of window = 15.5 m
Hence, Height of the Building is H = 4.51 + 1 + 15.5
⇒
H = 21.01 m
Problem 12
A driver having a definite reaction time (i.e., the interval between the perception of a signal to stop and the
application of brakes) is capable of stopping his car
over a distance of 30 m on seeing a red traffic signal
when the speed of the car is 72 kmh −1 and over a distance of 10 m when the speed of the car is 36 kmh −1 .
Find the distance over which he can stop the car if
it were running at a speed of 54 kmh −1 . Assume
that his reaction time and the deceleration of the car
remains same in all the three cases.
Solution
Let t0 is the reaction time, and a is the deceleration
During reaction time, the car travels at constant
speed. If u is speed of car, the distance traversed in
reaction time t0 is ut0 , therefore for stopping distance s , the distance traversed with deceleration a
is ( s − ut0 ) and final velocity = 0
CASE-1
Initial speed u1 = 72 kmh −1 = 20 ms −1
s1 = 30 m
11/28/2019 7:09:12 PM
4.85
Chapter 4: Kinematics I
⇒
0 = u12 − 2 a ( 30 − u1to )
⇒
u12 = 2 a ( 30 − 20u1to )
⇒
( 20 ) = 2a ( 30 − u1to ) …(1)
⇒
2
CASE-2
u2 = 36 kmh −1 = 10 ms −1 , s2 = 10 m
⇒
( 10 − u2to )
u22 = 2 a ( 10 − u2 to )
⇒
( 10 )2 = 2a ( 10 − 10to ) …(2)
0 = u22 − 2 a
⇒
Solving (1) and (2),
⇒ t0 = 0.5 s and a = 10 ms −2 {Retardation}
CASE-3
u3 = 54 kmh
−1
= 15 ms
−1
s2 = ( l1 − v1t ) + ( l2 − v2 t )
2
For s to be minimum,
0 = ( 15 ) − 2 × 10 × ( s3 − 15 × 0.5 )
⇒
s3 = 18.75 m
2
( )
ds
d 2
= 0 or
s =0
dt
dt
ds
= 2 ( l1 − v1t ) ( −v1 ) + 2 ( l2 − v2 t ) ( −v2 ) = 0
dt
⇒
2s
⇒
−l1v1 + v12 t − l2 v2 + v22 t = 0
⇒
l v +l v
t = 1 12 22 2
v1 + v2
Minimum s is given by
⎡
⎤
⎛ l v +l v ⎞
2
smin
= ⎢ l − v1 ⎜ 1 12 22 2 ⎟ ⎥
⎝ v1 + v2 ⎠ ⎥⎦
⎢⎣
⇒
2
smin
=
⇒
smin =
s3 = ?
⇒
2
2
⎡
⎛ l v +l v ⎞ ⎤
+ ⎢ l2 − v2 ⎜ 1 12 22 2 ⎟ ⎥
⎝ v1 + v2 ⎠ ⎥⎦
⎢⎣
2
( l1v2 − l2 v1 )2
v12 + v22
l1v2 − l2 v1
v12 + v22
Problem 14
Problem 13
Two particles A and B move with constant velocity
v1 and v2 along two mutually perpendicular straight
lines towards intersection point O . At moment t = 0
particles were located at distances l1 and l2 respectively from O . How soon will the distance between
particles be minimum and what is that minimum distance equal to?
Two bodies move in the same straight line at the
same instant of time from the same origin. The first
body moves with a constant velocity of 40 ms −1 and
second starts with a constant acceleration of 4 ms −2 .
Find the time t that elapses before the second catches
the first body. Find also the greater distance between
them prior to it and the time at which this occurs.
Solution
Solution
Let the separation between the particles be minimum
at time t . Then
Let the two bodies meet after a time t . The distance
travelled by both is the same
The distance travelled by first body is
1
v1t
A
s1 = 40 × t
The distance travelled by second body is
S
1
s2 = 0 + ( 4 ) t 2
2
Since s1 = s2
B
2
l1
l2
O
⇒
v2t
Since
2
OB = l2 − v2 t
2
AB = OB + OA
04_Kinematics 1_Part 3.indd 85
and
OA = l1 − v1t
and
⇒
1
( 4 ) t2
2
t = 20 s
40t =
2
11/28/2019 7:09:21 PM
4.86 JEE Advanced Physics: Mechanics – I
The distance between the two bodies goes on increasing as long as the velocity of the second body remains
less than 40 ms −1 . The distance between them will be
greatest prior to their meeting when the velocity of
the second body becomes 40 ms −1 . Let t1 be the time
when this happens. Then
40 = 0 + 4t1 {∵ v = u + at }
The time during which the motor cyclist moves with
constant speed is ( t − 10 ) s
The distance travelled at constant speed is
⇒
For Car: Maximum speed attained is
t1 = 10 s
So the distance between them will be greatest after
t1 = 10 s . Let x1 and x2 be the distances moved by
the two bodies in t1 = 10 s , then
⇒
1
2
x1 = ( 40 × 10 ) + ( 0 )( 10 )
2
x1 = 400 m
1
2
and x2 = 0 + ( 4 ) ( 10 ) 2
⇒ x2 = 200 m
Greatest distance between them is Δx = x1 − x2
⇒ Δx = 400 − 200 = 200 m
Problem 15
A motor cycle and a car start from rest at the same
place at the same time and travel in the same direction. The cycle accelerates uniformly at 1 ms −2 upto
a speed of 36 kmh −1 and the car at 0.5 ms −2 upto a
speed of 54 kmh −1 . Calculate the time and distance
at which the car overtakes the cycle.
x2 = 10 ( t − 10 ) = ( 10t − 100 ) m …(2)
Total distance x = x1 + x2
⇒
x = 50 + 10t − 100 = ( 10t − 50 ) m …(3)
v2 = 54 kmh–1 = 15 ms–1
Its acceleration is a2 = 0.5 ms–2. Let t2 be the time
taken to reach the maximum speed. Then
15 = 0 + ( 0.5 ) t2
15
= 30 s
0.5
The distance travelled before reaching the maximum
speed is
1
2
x3 = 0 + ( 0.5 )( 30 ) = 255 m …(4)
2
⇒ t2 =
The time for which the car is moving at constant
speed is ( t − 30 ) s
The distance travelled at constant speed
x 4 = 15 ( t − 30 ) = 15t − 450 …(5)
Total distance travelled is x ′ = x3 + x 4
⇒
x ′ = 225 + 15t − 450 = 15t − 225 …(6)
Equating equations (3) and (6)
10t − 50 = 15t − 225
Solution
⇒
5t = 175
When the car overtakes the motor cycle, the two have
travelled the same distance in the same time. Let the
time taken be t second while the total distance travelled be x metre
⇒
t = 35 s
For Motor Cycle: Maximum speed attained is
v1 = 36 kmh–1 = 10 ms–1
–2
Its acceleration is a1 = 1 ms . Let t1 be the time taken
to reach the maximum speed, then using v = u + at,
we get
10 = 0 + ( 1 ) ( t1 )
⇒
t1 = 10 s
The distance travelled before reaching the maximum
speed is
1
1
2
x1 = 0 + ( 1 ) t12 = ( 1 ) ( 10 ) = 50 m …(1)
2
2
04_Kinematics 1_Part 3.indd 86
Substituting this value of t either in equation (3) or in
equation (6), we get
x = x ′ = ( 10 )( 35 ) − 50 = 300 m
Problem 16
An airport shuttle train travels between two terminals that are 2.5 km apart. To maintain passenger
comfort, the acceleration of the train is limited to
±1.2 ms −2 and the jerk, or rate of change of acceleration, is limited to ±0.24 ms −3 . If the shuttle has a maximum speed of 30 kmh −1 , determine
(a) the shortest time for the shuttle to travel between
the two terminals,
(b) the corresponding average velocity of the shuttle.
11/28/2019 7:09:28 PM
4.87
Chapter 4: Kinematics I
From B to C, we have a = 1.2 − 0.24t
Solution
(a) Corresponding a-t graph is as shown in figure
⇒
1.2
E
O
–1.2
A
B
C
F
G
16
+ 1.2t − 0.12t 2
3
xBC
5
0
0
⎛ 16
2⎞
80
2
3
+ 0.6 ( 5 ) − 0.04 ( 5 ) = 36.67 m
3
⇒
xBC =
Now let tAB = tEF = t1 and tCD = t2
⇒
xOC = xDG = ( 5 + 8.1 + 36.67 ) = 49.77 m
According to the problem, maximum speed is
⇒
xOC = xDG ≅ 50 m
⇒
xOC + xDG = 100 m
In the figure tOA = tBC = tDE = tFG =
0
∫ dx = ∫ ⎜⎝ 3 + 1.2t − 0.12t ⎟⎠ dt
Also
1.2
=5s
0.24
16 3
v=
t(s)
D
t
∫ dv = ∫ ( 1.2 − 0.24t ) dt
⇒
a(ms–2)
v
vmax = vC = 30 kmhr −1 =
25
ms −1
3
Since vC = Area under a-t graph from O to C,
⇒
25 1
= × 1.2 ( t1 + t1 + 10 )
3 2
⇒
t1 =
35
s
18
The remaining distance ( 2500 − 100 ) m or
2400 m must have been travelled with a constant
25
speed of
ms −1 . Hence
3
2400
= 288 s
t2 =
25 3
(a) Total time
35
⎛
⎞
T = 2⎜ 5 +
+ 5 ⎟ + 288 = 312 s = 5.2 minute
⎝
⎠
18
From O to A , we have, a = 0.24t
v
⇒
∫
0
∫
0.24tdt
0
⇒
v = 0.12t 2
⇒
vA = ( 0.12 )( 5 ) = 3 ms −1
2
xOA
Again,
∫
Problem 17
5
An open lift is moving upwards with velocity
10 ms −1 . It has an upward acceleration of 2 ms −2 .
A ball is projected upwards with velocity 20 ms −1
relative to ground. Find
∫
dx = 0.12 t 2 dt
0
⇒
(b) Average velocity,
2.5
vav =
kmhr −1 = 28.8 kmhr −1
5.2
60
t
dv =
0
xOA = 5 m
(a) time when ball again meets the lift.
(b) displacement of lift and ball at that instant.
(c) distance travelled by the ball upto that instant.
From A to B , we have
a = 1.2 ms −2 = constant
⇒
⎛ 35 ⎞ 16
vB = vA + atAB = 3 + ( 1.2 ) ⎜ ⎟ =
ms −1
⎝ 18 ⎠
3
1
2
x AB = vA tAB + a ( tAB )
2
⇒
Solution
(a) At the time when ball again meets the lift,
2
⎛ 35 ⎞ 1
⎛ 35 ⎞
x AB = ( 3 ) ⎜ ⎟ + ( 1.2 ) ⎜ ⎟ ≅ 8.1 m
⎝ 18 ⎠ 2
⎝ 18 ⎠
04_Kinematics 1_Part 3.indd 87
Take g = 10 ms -2
sL = sB
⇒
10t +
1
1
× 2 × t 2 = 20t − × 10t 2
2
2
11/28/2019 7:09:36 PM
4.88 JEE Advanced Physics: Mechanics – I
2 ms–2
10 ms–1
20 ms–1
Ball (B)
LIFT (L)
10 ms–2
Solving, we get
5
s
3
5
So, ball will again meet the lift after s
3
(b) At this instant
t = 0 and t =
2
175
⎛ 5⎞
⎛ 5⎞ 1
sL = sB = 10 ⎜ ⎟ + ( 2 ) ⎜ ⎟ =
m = 19.4 m
⎝ 3⎠
⎝ 3⎠ 2
9
(c) For the ball u and g are antiparallel. Therefore
we will first find t0 , the time when its velocity
becomes zero.
Since v = u + at
⇒
0 = u + ( − g ) t0
⇒
t0 =
Given, vsr = 3 kmhr −1 , vr = 2 kmhr −1
and vw = walking speed = 5 kmhr −1
vs = vsr + vr
⇒
and ( vs )y = vsr cos θ = 3 cos θ
Time taken to reach the other side
0.5
1
l
t1 =
=
=
θ
cos
θ
3
cos
6
v
( s )y
Horizontal drift x is given by
1
tan θ
1
=
−
6 cos θ 3 cos θ
2
Time to travel this distance by walking
1
tan θ
x
=
−
t2 =
…(1)
10
vw 15 cos θ
x = ( vs )x t1 = ( 2 − 3 sin θ )
Total time t = t1 + t2 =
d = 19.4 m
7
sec θ tan θ − sec 2 θ = 0
3
3
⇒ sin θ =
7
From equation (2), we get
Problem 18
A man can row a boat in still water at 3 kmh −1 . He
can walk at a speed of 5 kmh −1 on the shore. The
water in the river flows at 2 kmh −1 . If the man rows
across the river and walks along the shore to reach the
opposite point on the river bank, find the direction in
which he should row the boat so that he could reach
the opposite shore in the least possible time. Also calculate this time. The width of the river is 500 m .
Solution
Suppose the boatman rows with velocity vsr in the
direction shown in figure
y
vsr
l = 0.5 km
θ
vr
dt
=0
dθ
⇒
tmin =
x
1 ⎛ 7
⎞
− tan θ ⎟ …(2)
⎜
⎠
10 ⎝ 3 cos θ
For time to be minimum, we have
u 20
=
=2s
g 10
⎛ 5 ⎞
Since t ⎜ = s ⎟ < t0 , so the distance and displace⎝ 3 ⎠
ment are equal i.e.,
04_Kinematics 1_Part 3.indd 88
( vs )x = vr − vsr sin θ = 2 − 3 sin θ
⇒
1 ⎛7
7
3 ⎞
×
−
⎟
10 ⎜⎝ 3
40
40 ⎠
tmin = 0.21 hour = 756 s
Problem 19
The velocity of water current in a river changes with
distance along the perpendicular to the river according to the function
d
⎡ 2u0
for 0 ≤ y ≤
⎢ d y
2
u=⎢
u
2
d
0
⎢
d − y ) for
≤y≤d
⎢⎣ d (
2
where u0 is velocity at the mid-point of the river and
d is width of the river. A boat travels from a point O
on one bank of the river to the opposite bank and its
steering angle is adjusted to keep its relative velocity perpendicular to the river current. Calculate the
time in which the boat will reach the other bank. The
velocity of the boat in still water is u0 .
11/28/2019 7:09:42 PM
Chapter 4: Kinematics I
Solution
Denoting the river current by R and the boat by B.
Let the boat steer making an angle θ with the river
flow. Then
uR = uiˆ , uB = u0 ( cos θ iˆ + sin θ iˆ )
4.89
­ anner that it is always perpendicular to current and
m
the speed of the boat in still water is u . Find the distance through which the boat crossing the river will
be carried away by the current if the width of river is
l . Also determine the trajectory of boat.
iˆ
C
uBR = uB − uR = ( u0 cos θ − u ) iˆ + u0 sin θ ˆj
iˆ
B
As the boat is steered perpendicular to flow, so we
have
( uBR )x = 0
A
⇒
u0 cos θ − u = 0
⇒
cos θ =
⇒
u2
uBR = u0 sin θ = μ0 1 − 2 = u02 − u2
u0
⇒
dy
= u02 − u2
dt
⇒
∫
dt =
t=
0
⇒
t=
d
2u0
2
0
2
d
dy
4u 2
u02 − 20 y 2
+
∫
d
d
2
dy
∫ ⎛ d⎞
0
2
⇒
d
2
+
2
⎜⎝ ⎟⎠ − y
2
2v0
…(2)
l
Since y = ut …(3)
So, equation (1) becomes
⇒
dy
u02 −
d
2u0
4u02
2
d− y)
2 (
d
d
∫ ⎛ d⎞
d
2
2
⎜⎝ ⎟⎠ − ( d − y )
2
⇒
2
dvx
2v u
= ax = 0 = constant
dt
l
1 2
ax t
2
d
Since x = ux t +
2
⇒
1 ⎛ 2v u ⎞
x = 0 + ⎜ 0 ⎟ t2
2⎝ l ⎠
⇒
⎛ v u⎞
x = ⎜ 0 ⎟ t 2 …(4)
⎝ l ⎠
d ⎛π
d ⎡
⎞
⎛ π ⎞ ⎤ πd
0−⎜ ⎟ ⎥ =
⎜ − 0 ⎟⎠ +
⎝ 2 ⎠ ⎦ 2u0
2u0 ⎝ 2
2u0 ⎢⎣
Problem 20
The current velocity of a river grows in proportion to
the distance from its bank and reaches the maximum
value v0 in the middle. Near the banks the velocity
is zero. A boat is moving along the river in such a
04_Kinematics 1_Part 3.indd 89
k=
⎛ 2v ⎞
vx = ⎜ 0 ⎟ ut
⎝ l ⎠
dy
d ⎡ −1 ⎛ 2 y ⎞ ⎤ 2
d ⎡ −1 ⎛ 2(d − y ) ⎞ ⎤
t=
⎟⎥ +
⎟⎥ d
⎢ sin ⎜
⎢ sin ⎜
2u0 ⎣
d
⎝ d ⎠ ⎦ 0 2u0 ⎣
⎝
⎠⎦
t=
l
2
From (1), we get
⎛ l⎞
v0 = k ⎜ ⎟
⎝ 2⎠
dy
d
⇒
Also, we have been given in the problem, that
vx = v0 when y =
∫ u −u
∫
Let the river flow or the current flow be along x-axis.
At any instant t , let the boat be at a distance y from
the bank, then
vx = k y …(1)
u
u0
d
2
⇒
Solution
l
y
11/28/2019 7:09:51 PM
4.90 JEE Advanced Physics: Mechanics – I
From (3), t =
y
u
Now, when y =
l
, we have, from (5)
2
⇒
⎛ v u⎞
x = ⎜ 02 ⎟ y2
⎝ lu ⎠
l 2 ⎛ lu ⎞
x
=
4 ⎜⎝ v0 ⎟⎠
⇒
⎛v ⎞
x = ⎜ 0 ⎟ y2
⎝ lu ⎠
⇒
⎛ lu ⎞
y 2 = ⎜ ⎟ x …(5)
⎝ v0 ⎠
lv0
4u
Hence total drift is
This happens to be the equation of a parabola
04_Kinematics 1_Part 3.indd 90
⇒
x=
Drift = 2x =
⇒ Drift =
2lv0
4u
lv0
2u
11/28/2019 7:09:53 PM
Chapter 4: Kinematics I
4.91
Practice Exercises
Single Correct Choice Type Questions
This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
1.
2.
3.
4.
A particle has an initial velocity of 9 ms–1 due east and
a constant acceleration of 2 ms–2 due west. The distance covered by the particle in the fifth second of its
motion is
(A) 0 m
(B) 0.5 m
(C) 2 m
(D) 20 m
(B) 1 ms −2
(C) −1 ms −2
(D) −2 ms −2
(B)
(C) 144 kmh −1
(D) 174 kmh −1
Ball A is dropped from the top of a tower of height H.
At the same instant ball B is thrown vertically upwards
from the ground. When the balls collide, they are moving in opposite directions and the speed of A is twice
the speed of B. The height from the ground where the
collision happens is
H
H
(B)
(A)
3
4
(C)
5.
6.
72 kmh −1
2H
5
(D)
2H
3
Wind is blowing at a harbour with a speed of 72 kmh −1
and the flag on the mast of a boat anchored at harbour
flutters along the North-East direction. If the boat
starts moving at a speed of 51 kmh −1 due North, the
direction of the flag is (in approximation)
(A) towards east
(B) towards west
(C) towards north
(D) towards south
Two bodies begin to fall freely from the same height
but the second falls T second after the first. The time
(after which the first body begins to fall) when the distance between the bodies equals L is
04_Kinematics 1_Part 3.indd 91
(B)
T
L
+
2 gT
(C)
L
gT
(D) T +
2L
gT
A particle is dropped from point A at a certain height
from ground. It falls freely and passes through three
points B, C and D with BC = CD. The time taken by the
particle to move from B to C is 2 seconds and from C
to D is 1 second. The time taken to move from A to B is
(A) 0.25 s
(B) 0.5 s
(C) 0.75 s
(D) 1.5 s
8.
A juggler maintains four balls in motion, making each
of them to rise a height of 20 m from his hand. The
time interval that he should maintain, for the proper
distance between them g = 10 ms-2 is
A racing car travelling at a constant speed has to pass
through a horizontal turn where the radius of curvature of the road is 200 m. If the normal acceleration
of the car cannot exceed 0.8 g where g = 10 ms −2 , the
maximum speed of the car without sliding can be
(A) 36 kmh −1
T
2
7.
The velocity of a car travelling on a straight road is
given by the equation v = 9 + 8t − t 2 where v is in
metre per second and t in second. The instantaneous
acceleration when t = 5 s is
(A) 2 ms −2
(A)
(A) 0.5 s
(C) 2 s
9.
(B) 1 s
(D) 3 s
A particle starts from rest at time t = 0 and undergoes
acceleration a as shown. The velocity as function of
time during the interval 0 to 4 second is indicated in
a(ms–2)
3
2
1
0
–1
–2
–3
1 2
3 4
(A) v(ms–1)
5
t(s)
(B) v(ms–1)
t(s)
t(s)
(C) v(ms–1)
(D) v(ms–1)
t(s)
t(s)
11/28/2019 7:09:57 PM
4.92 JEE Advanced Physics: Mechanics – I
10. The slopes of the windscreen of two motor cars are
β1 = 30° and β2 = 15° respectively. The cars have
velocities v1 and v2 in the horizontal direction. If
the hailstones appear to the drivers to be bounced by
the windscreen of their respective cars in the vertical
v
direction then 1 (assuming that hailstones were fallv2
ing on the cars vertically) is
(A) 3
(B) 1
(C)
1
3
(D)
(A)
t1 + t2
2
(B)
t1 + t2
(C)
t1t2
(D)
t1t2
t1 + t2
(B)
s
t
(C)
t
(D)
s
(C) v = 2 gt
(D) v = 16 gt
v(ms–1)
10
s(m)
10
(A) a(ms–2)
(B) a(ms–2)
10
10
v(ms–1)
10
(C) a(ms–2)
(D) a(ms–2)
10
10
10
v(ms–1)
10
v(ms–1)
v(ms–1)
16. Passengers in the jet transport A flying east at a speed
of 800 kmh −1 observe a second jet plane B that passes
under the transport in horizontal flight. Although the
nose of B is pointed in the 45° north east direction,
plane B appears to the passengers in A to be moving
away from the transport at the 60° angle as shown.
The true velocity of B is
s
60°
t
B
t
A
45°
13. Wind is blowing from the south at 10 ms −1 but to
a cyclist it appears to be blowing from the east at
10 ms −1 . The cyclist has a velocity
(A) 10i − 10 j
(B) 10i + 10 j
(C) −10i + 10 j
(D) −10i − 10 j
14. If a particle takes t second less and acquires a velocity
of v ms–1 more in falling through the same distance on
two planets where the accelerations due to gravity are
2 g and 8 g respectively then
04_Kinematics 1_Part 3.indd 92
v = 5 gt
15. Velocity versus displacement graph of a particle moving in a straight line is shown in figure. Corresponding
acceleration versus velocity graph will be
10
12. One stone is dropped from a tower from rest and
simultaneously another stone is projected vertically
upwards from the tower with some initial velocity.
The graph of the distance, s between the two stones
varies with time ( t ) as (before either stone hits the
ground)
s
(B)
1
9
11. A person walks up a stationary escalator in time t1. If
he remains stationary on the escalator, then he reaches
up in time t2 . The time it would take him to walk up
the moving escalator is
(A)
(A) v = 4 gt
x
(A) 586 kmh −1
(B)
400 2 kmh −1
(C) 717 kmh −1
(D) 400 kmh −1
17. A body falling freely from a tower of height h covers
7
a distance of
h during the last second of its motion.
16
Then the height of tower is (Take g = 10 ms −2 )
(A) 60 m
(B) 70 m
(C) 80 m
(D) 90 m
11/28/2019 7:10:03 PM
4.93
Chapter 4: Kinematics I
18. The figure shows the displacement-time graph (a
parabola) of a body. This graph indicates that the initial velocity, in ms −1 and acceleration, in ms −2 respectively are
(A)
m
( v2 − v1 )
3P
(C)
m 3
v2 − v13
3P
s(m)
40
20
1
(A) 80, 40
(C) 80, 32
2
3
4
5
t(s)
(B) 40, 80
(D) 32, 16
19. Acceleration of a particle is a for a time t . It is folt
lowed immediately by a retardation of a for time .
2
Consider this as one cycle. If initial velocity of particle
is zero, then the displacement of the particle after n
such cycles in succession is
(C)
n ( 3n + 4 ) 2
at
8
( n + n + 1)
2
4
at 2
(B)
n( n + 1) 2
at
2
(D) nat 2
20. A stone tied to a string of length L is whirled in a
vertical circle with the other end of the string at the
centre. At a certain instant of time, the stone is at the
lowest position and has a speed u . The magnitude of
the change in velocity as it reaches a position where
the string is horizontal is
(A)
u2 − 2 gL
(B)
2gL
(C)
u2 − gL
(D)
2 u2 − gL
(
)
21. For a particle moving rectilinearly the displacement x
depends on time t as x = at 3 + bt 2 + ct + d. The ratio of
its initial acceleration to its initial velocity depends
(B) only on b and c
(A) only on a and b
(C) only on a and c
(D) only on a
22. A self-propelled vehicle of mass m whose engine
P
mv
(assume that there is no friction). In order to increase
its velocity from v1 to v2 , the distance it has to travel
will be
delivers constant power P has an acceleration a =
04_Kinematics 1_Part 3.indd 93
)
3P 2
v2 − v12
m
dv
= a − bv
dt
where a and b are constants. The velocity at any time
t is
60
(A)
(
(D)
medium is described by the equation
80
0
)
)
m 2
v2 − v12
3P
23. The motion of a body falling from rest in a resisting
a
100
(
(
(B)
(A) vt =
a(
1 − e − bt )
b
(B)
vt =
b − bt
e
a
(C) vt =
a(
1 + e − bt )
b
(D) vt =
b bt
e
a
24. A parachutist drops freely from an airplane for 10 s
before the parachute opens. He then descends with a
uniform retardation of 2.5 ms–2. If he bails out of the
plane at a height of 2495 m and g is 10 ms–2, his velocity on reaching the ground will be
(A) 5 ms −1
(B) 10 ms −1
(C) 15 ms −1
(D) 20 ms −1
25. The nucleus of a helium atom travels along the axis of
a straight hollow tube 4 m long. The tube is part of a
particle accelerator. If the particle enters the tube with
a speed of 1000 ms −1 and leaves at 9000 ms −1 , and
assuming that the acceleration is uniform, the time the
particle remains inside the tube is
(A) 4 × 10 −3 s
(B)
8 × 10 −4 s
(C) 1 × 10 −4 s
(D) 8 × 10 −3 s
26. Two cars A and B start off to race with velocities
8 ms −1 and 4 ms −1 and travel in straight line with
uniform accelerations 2 ms −2 and 4 ms −2 respectively. If they reach the final point at the same instant,
then the length of the path is
(A) 24 m
(B) 32 m
(C) 48 m
(D) 16 m
x
and
27. A particle is moving in x -y plane with y =
2
vx = 4 − 2t . The displacement versus time graph of the
particle would be
(A)
s
(B)
t
s
t
11/28/2019 7:10:12 PM
4.94 JEE Advanced Physics: Mechanics – I
(C)
s
(D)
s
t
t
28. A ball is thrown vertically upwards. It was observed at
a height h twice with a time interval Δt . The initial
velocity of the ball is
2
(A)
8 gh + g 2 ( Δt )
(C)
1
2
8 gh + g 2 ( Δt )
2
2
(B)
⎛ g Δt ⎞
8 gh + ⎜
⎝ 2 ⎟⎠
(D)
8 gh + 4 g 2 ( Δt )
2
29. A point moves in x -y plane according to the law
x = 5 sin ( 6t ) and y = 5 ( 1 − cos ( 6t ) ) , where x and y
are in metre. The distance traversed by the particle in
t = 4 s is
(A) 24 m
(B)
(C) 96 m
(D) 120 m
48 m
30. A particle moving with uniform acceleration along a
straight line covers distances a and b in successive
intervals of p and q second. The acceleration of the
particle is
(A)
(C)
pq ( p + q )
2 ( bp − aq )
2 ( ap − bq )
pq ( p + q )
(B)
2 ( aq − bp )
(D)
2 ( bp − aq )
pq ( p + q )
pq ( p + q )
31. The motion of a particle is defined by x = a cos ( ωt )
and y = a sin ( ωt ) . The acceleration of the particle is
(A) aω
(B)
a 2ω
(C) aω 2
(D)
aω 2
2
32. The figure shows the acceleration versus time graph
of a train. If it starts from rest, the distance it travels
before it comes to rest is
(A) 30 metre
(B)
(C) 13 metre
(D) 40 metre
33. A car A is going north east at 80 kmh–1 and another
car B is going south east with a velocity of 60 kmh −1 .
The velocity of A relative to B makes an angle with the
north equal to
⎛ 2⎞
(A) tan −1 ⎜ ⎟
⎝ 7⎠
(B)
(C) tan −1 ( 7 )
⎛ 1⎞
(D) tan −1 ⎜ ⎟
⎝ 7⎠
1
0
–1
–1.5
04_Kinematics 1_Part 3.indd 94
1
2
3 4
t(s)
⎛ 7⎞
tan −1 ⎜ ⎟
⎝ 2⎠
34. A particle is moving along a circular path of radius
3 metre in such a way that the distance travelled meat2 t3
+ .
2 3
The acceleration of the particle when t = 2 second is
sured along the circumference is given by s =
(A) 1.3 ms −2
(B)
(C) 10 ms −2
(D) 13 ms −2
3 ms −2
35. A lead ball is dropped into a lake from the diving board
5 m above the water. In hits the water with a certain
velocity and then sinks to the bottom with this same
constant velocity. It reaches the bottom 5 s after it is
(
dropped. The depth of the lake is take g = 10 ms −2
(A) 10 m
(B)
(C) 25 m
(D) 40 m
)
20 m
36. A balloon starts rising from the ground with an acceleration of 1.25 ms −2 . After 8 s, a stone is released from
the balloon. The stone will
(A) cover a distance of 40 m
(B) have a displacement of 50 m
(C) reach the ground in 4 s
(D) begin to move down after being released
37. The velocity acquired by a body when it falls through
a height h is v . If it further falls through a height x
( x h ) , the increase in velocity is approximately
(A)
vx
2h
(B)
2v
xh
(C)
2vx
h
(D)
v
2xh
a(ms–2)
2
26 metre
38. An armoured car 2 m long and 3 m wide is moving
at 10 ms −1 when a bullet hits it in a direction making
3
an angle tan −1 ⎛⎜ ⎞⎟ with the length of the car as seen
⎝ 4⎠
11/28/2019 7:10:22 PM
Chapter 4: Kinematics I
4.95
by a stationary observer. The bullet enters one edge of
43. The position of a particle is given by r = a cos(ωt)i + a sin(ωt)j + btk
the car at the corner and passes out at the diagonally
2π
r = a cos(ωt)i + a sin(ωt)j + btk where ω =
and T is time period
opposite corner. Neglecting any interaction between
T
the car and the bullet, the time for the bullet to cross
for one revolution of the particle following a helical
the car is
path. The distance moved by the particle in one full
(A) 0.20 s
(B) 0.15 s
turn of the helix is
(C) 0.10 s
(D) 0.50 s
4π 2
2π 2 2
a + b 2ω 2
(B)
a ω + b2
(A)
39. A time-velocity graph of two vehicles P and Q startω
ω
ing from rest at the same time is given in the figure.
2π 2
4π 2 2
The statement that can be deduced correctly from the
a + b 2ω 2
(D)
a ω + b2
(C)
ω
ω
graph is
44. The graph in the figure shows the velocity of a body
plotted as a function of time. The distance covered by
the body in the first 12 s is
V
Q
P
v(ms–1)
50
40
30
t
(A) velocity of Q is greater than that of P
(B)acceleration of P is increasing at a slower rate
than that of Q
(C) acceleration of Q is greater than that of P
(D) acceleration of P is greater than that of Q
40. A stone takes time t to fall through a height h. The
increment in time when it falls further through a distance x ( x h ) is
(A)
xth
2
(B)
xt
2
(C)
xt
2h
(D)
2h
xt
20
10
0
(A) 98 metre
(B)
(C) 973 metre
(D) 1800 metre
450 metre
(A) 10 ms
(B) 15 ms
−1
(D) 25 ms −1
(C) 20 ms
04_Kinematics 1_Part 3.indd 95
6
t(s)
8 10 12
(B) 240 m
(D) 500 m
(C)
v
a
v
2a
(B)
2v
a
(D)
2v
a
46. A graph between the square of the velocity of a particle
and the distance s moved by the particle is shown in
the figure. The acceleration of the particle in kilometre
per hour square is
42. In a 100 metre race, a runner accelerates uniformly
from the start to his maximum velocity in a distance of
4 m and runs the remaining distance at that velocity.
If he finishes the race in 10.4 second , then his maximum velocity was
−1
4
45. A horizontal wind is blowing with a velocity v
towards north-east. A man starts running towards
north with acceleration a . The time after which man
will feel the wind blowing towards east is
(A)
41. The speed of an aeroplane at the instant it lands on
a runway is 60 ms −1 . If the deceleration of the aeroplane is given as a = −0.6 − 0.001 v 2 , the distance that
it covers on the runway before coming to a stop is
2
(A) 360 m
(C) 285 m
v 2 (in km/hr)2
O
3600
900
O
s (in km) 0.6
−1
(A) 2250
(C) –2250
(B) 225
(D) –225
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4.96 JEE Advanced Physics: Mechanics – I
(A) during this time it travels 52.5 metre .
(B) its acceleration is 5 ms −2 .
(C)
its acceleration is greatest at the beginning
because it is going fastest at that time.
(D)
the distance travelled during an interval of
3 second cannot be calculated from the given
data.
48. A car covers the first half of the distance between two
places at a speed of 40 kmh −1 and the second half at
60 kmh −1 . The average speed of the car is
(A) 50 kmh
−1
(B)
(C) 35 kmh
−1
(D) 48 kmh −1
42 kmh
−1
49. Rain, pouring down at an angle α with the vertical
has a speed of 10 ms −1 . A girl runs against the rain
with a speed of 8 ms −1 and sees that the rain makes an
angle b with the vertical, then relation between α and
b is
8 + 10 sin α
8 + 10 sin β
(B) tan β =
(A) tan α =
10 cos α
10 cos β
(C) tan α = tan β
(D) tan α = cot β
50. A smooth square platform ABCD is moving towards
right with a uniform speed v . A particle is projected
from A with speed 2v making an angle θ with AD
so that it strikes the point B . Then θ equals
B
C
v
2v
θ
A
D
(A) 30°
(B)
(C) 60°
(D) 90°
45°
51. An express elevator can accelerate or decelerate
with values whose magnitudes are limited to 0.4 g .
The elevator attains a maximum vertical speed of
400 metre per minute . The minimum time required
by the elevator to start from rest from the 10th floor
and to stop at the 30th floor, a distance 100 m apart is
(A) 1.67 s
(B) 16.7 s
(C) 167 s
(D) 1670 s
52. A swimmer crosses a flowing stream of width d to
and fro in time t1 . The time taken to cover the same
04_Kinematics 1_Part 3.indd 96
distance up and down the stream is t2 . If t3 is the time
the swimmer would take to swim a distance 2d in still
water, then
(B) t32 = t1t2
(A) t3 = t1 + t2
(C) t22 = t1t3
(D) t12 = t2t3
1
of its velocity in passing through a
20
plank. The least number of planks required to stop the
bullet is
(A) 10
(B) 11
(C) 12
(D) 23
53. A bullet loses
54. A motorboat going down stream overcame a raft at a
point A. 60 minute later it turned back and after some
time passed the raft at a distance of 6 km from the
point A. Assuming the duty of the engine to be constant, the flow velocity is
(B) 4 kmh–1
(A) 3 kmh–1
–1
(C) 5 kmh
(D) 6 kmh–1
55. A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves
horizontally with a speed of 2 ms −1 . The angle α with
the vertical at which the wind must screen be placed so
that the rain drops, falling vertically downwards with
velocity 6 ms −1 , strike the wind screen normally is
⎛ 1⎞
(A) sin −1 ⎜ ⎟
⎝ 3⎠
(B)
⎛ 1⎞
(C) tan −1 ⎜ ⎟
⎝ 3⎠
(D) tan −1 ( 3 )
cos −1 ( 3 )
56. A train travelling at 72 kmh −1 is checked by track
repairs. It retards uniformly for 200 m covering next
400 m at constant speed and accelerates to 72 kmh −1
in a further distance of 600 m. If the time at constant
lower speed is equal to the sum of the times taken in
retarding and accelerating, the total time taken is
(A) 140 s
(B) 160 s
(C) 120 s
(D) 100 s
57. For an airplane to take-off it accelerates according to
the graph shown and takes 12 s to take-off from the
rest position. The distance travelled by the airplane is
Acceleration
m/s2
47. The speed of a body moving in a straight line changes
from 25 ms −1 to 10 ms −1 in 3 s at a constant rate.
5
O
A
B
6
t (in s)
12
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Chapter 4: Kinematics I
(A) 21 m
(C) 2100 m
(B) 210 m
(D) 120 m
58. A body with constant acceleration travels 2 metre in
the first 2 second and 2.2 m in the next 4 second .
The velocity at the end of the seventh second from the
start shall be
(A) 0.1 ms −1
(B)
0.2 ms −1
(C) 0.5 ms −1
(D) 1 ms −1
59. A particle is released from rest from a tower of height
3h . The ratio of times to fall equal heights h , i.e.,
t1 : t2 : t3 is
(A) 3 : 2 : 1
(B)
(C) 1 : ( 2 − 1 ) : ( 3 − 2 )
(D) 9 : 4 : 1
60. Two particles A and B are thrown up simultaneously
from the edge of a cliff with initial speeds v and 2v.
Assuming that the particle A comes to rest immediately after striking the ground, the variation in relative
position of the particle B with respect to the particle
A with time, till both the stones strike the ground is
plotted. This variation plot is
(A) only linear
(B) only parabolic
(C) first parabolic then linear
(D) first linear then parabolic
61. A dog is chasing a cat who is running along a straight
line at constant speed u . The dog moves with a
constant speed v , always heading towards the cat.
Initially i.e. at t = 0 , the velocities of dog and cat are
perpendicular and the initial perpendicular distance
between them is l . The dog catches the cat at
(A) t =
lv
for v > u
v − u2
t=
lv
for u > v
u − v2
(B)
(C) t =
(D) t =
2
2
lv 2
2
v − u2
lv 2
u2 − v 2
for v > u
for u > v
62. The acceleration is constant when the relationship
between the
(A)position coordinate s and the square of the velocity v is linear
(B)position coordinate s and the velocity v is linear
04_Kinematics 1_Part 3.indd 97
(C)
position coordinate and the reciprocal of the
velocity v is linear
(D)square of the position coordinate s and the velocity v is linear
63. In travelling a distance of 3 kilometre between points
A and D , a car is driven at 100 kmh −1 from A to
B for t second and at 60 kmh -1 from C to D for t
second. If the brakes are applied for 4 second between
B and C to give the car a uniform deceleration, the
value of t is
100 kmh–1
A
3 : 2 :1
4.97
(A) 75.5 second
(C) 65.5 second
60 kmh–1
B C
3 km
D
(B) 45.5 second
(D) 56.5 second
64. A car is travelling on a straight road. The maximum
velocity the car can attain is 24 ms −1 . The maximum
acceleration and deceleration it can attain are 1 ms −2
and 4 ms −2 respectively. The shortest time the car
takes to start from rest and come to rest in a distance of
200 metre is
(A) 22.4 second
(B) 33.6 second
(C) 11.2 second
(D) 5.6 second
65. A person walks up a stalled escalator in 90 s . When
standing on the same escalator, now moving, he is carried in 60 s . The time it would take him to walk up the
moving escalator will be
(B) 18 s
(A) 36 s
(C) 72 s
(D) 27 s
66. Two particles, A and B move with constant velocities
vA and vB . Initially their radius vectors are rA and
rB . For the particles to collide the four vectors must be
interrelated as
(A) vA − vB = rA − rB
(B) vA = vB and rA = rB
r +r
v +v
(C) A B = A B
vA + vB
rA + rB
v −v
r −r
(D) B A = A B
vB − vA
rA − rB
67. The cone falling with a speed v0 strikes and penetrates the block of packing material. The acceleration
of the cone after impact is a = g − cx 2 , where c is a
positive constant and x is the penetration distance. If
the maximum penetration depth is xm . Then c equals
11/28/2019 7:10:44 PM
4.98 JEE Advanced Physics: Mechanics – I
v0
(B) s
s
x
1
2
(A) 2 gxm + v0
2
xm
(C)
6 gxm − 3v02
3
2xm
(B)
2 gxm − v02
2
xm
(D)
6 gxm + 3v02
3
2xm
)
1
2
3
(B) 14 ms −1
(C) 14 2 ms −1
(D) 17 ms −1
varies with time as t − 3 ms −1 where t is in second.
The total distance moved by the car is
(A) 3 m
(B) 6 m
(C) 9 m
(D) 12 m
2
2
1
1
–1
2 3
4
t(s)
0
1
–1
–2
2 3
t(s)
4
–2
s
(A)
s
2
2
1
0
–1
–2
04_Kinematics 1_Part 3.indd 98
1
1
2 3
4
t
0
–1
–2
1
4
t
1
2
3
4
t
s
t
t
a(ms–2)
5
O
1
0
t
A
B
4
8
2
71. The V -t curves for two different particle motions are
shown. The corresponding displacement time curves,
taking S = 0 when t = 0 for both instances will be
V ms–1
2 3
72. The acceleration of a train between two stations
2 kilometre apart is shown in the figure. The maximum speed of the train is
70. A car moves rectilinearly for 6 s with a velocity that
V ms–1
4
(D) s
)
(A) 20 ms −1
1
t
s
69. A particle has an initial velocity u = 6i + 8 j ms −1 and
an acceleration of a = 0.8i + 0.6 j ms −2 . Its speed after
10 s is
(
4
(C) s
68. Water drops fall at regular intervals from a roof. At
an instant when a drop is about to leave the roof, the
separations between 3 successive drops below the roof
are in the ratio
(A) 1: 2 : 3
(B) 1 : 4 : 9
(C) 1 : 3 : 5
(D) 1 : 5 : 13
(
2 3
2 3
4
t
–2
–5
t2
t3
t(s)
C
D
(A) 30 ms −1
(B)
60 ms −1
(C) 90 ms −1
(D) 120 ms −1
73. The acceleration is constant when the relationship
between
(A)the square of the position coordinate x and the
velocity v is linear
(B)the position coordinate x and the reciprocal of
the velocity is linear
(C)the position coordinate x and velocity v is linear
(D)the position coordinate x and the square of the
velocity v is linear
74. A particle moves along a horizontal straight line with a
velocity-time relationship as shown in figure. The total
distance moved by the particle is
11/28/2019 7:10:48 PM
Chapter 4: Kinematics I
V(ms–1)
8
2
0
5
t(s)
12
(B) 13 m
(D) 2.6 m
75. A train moves from one station to another in 2 hour,
and its speed during the motion is shown in the graph.
The maximum acceleration during the journey is
Speed (kmh–1)
D
60
40
20
0 A
B
C
1
E
2
t(h)
(A) 80 kmh −2
(B) 160 kmh −2
(C) 40 kmh −2
(D) 60 kmh −2
76. The wind appears to blow from the north to a man
moving in the north-east direction. When he doubles
his velocity the wind appears to move in the direction cot −1 ( 2 ) east of north. The actual direction of the
wind is
v
(A)
2v towards east
(B)
towards west
2
(C)
2v towards west
(D)
v
towards east
2
77. A man drives a car from Y towards X at speed
60 kmh −1 . A car leaves station X for station Y every
10 min. The distance between X and Y is 60 km .
The car travels at speed 60 kmh −1 . A man drives a car
from Y towards X at speed 60 kmh −1 . If he starts at
the moment when first car leaves station X . The number of cars he would meet on route is
(A) 5
(B) 7
(C) 10
(D) 20
78. A body falls from rest, in the last second of its fall,
it covers half of the total distance. If g is 9.8 ms −2 ,
then the total time of its fall is (in second)
04_Kinematics 1_Part 3.indd 99
2+ 2
(A) 2
(B)
(C) 2 − 2
(D) 2 ± 2
79. The motion of a body falling from rest in a resisting
dv
= A − Bv
medium is described by the equation
dt
where A and B are constants. The velocity at any
time t is
(A) A ( 1 − e Bt )
(A) 39 m
(C) 26 m
4.99
(C)
A(
1 − e − Bt )
B
(B)
ABte − t
(D) AB2 ( 1 − e − Bt )
80. In a car race, car A takes t0 time less to finish than car
B and passes the finishing point with a velocity v0
more than car B . The cars start from rest and travel
with constant accelerations a1 and a2 . Then the ratio
v0
is equal to
t0
(A)
a1 + a2
2
(B)
a22
a1
(C)
a12
a2
(D)
a1a2
81. A 2 m wide car is moving with a uniform speed of
8 ms −1 along the edge of a straight horizontal road.
A pedestrian starts to cross the road with a speed v
when the car is 12 m away from him. The minimum
value of v for the pedestrian to cross the road safely is
(A)
4
ms −1
3
(B)
3
ms −1
4
(C)
1
ms −1
2
(D)
1
ms −1
6
82. In PROBLEM 81, if q is the angle made by the velocity
of the pedestrian with the road then
(A) tan θ = 2
(B)
tan θ =
1
6
(C) tan θ = 6
(D) tan θ =
1
2
83. In PROBLEM 81, the time to cross the moving vehicle
safely is
(A)
24
s
37
(B)
37
s
24
(C)
24
s
49
(D)
3
s
2
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4.100 JEE Advanced Physics: Mechanics – I
84. A proton in a cyclotron moves in a circle of radius
0.8 metre at a speed of 107 ms −1 . The acceleration of
the proton and acceleration due to gravity have a ratio
of approximately
(A) 1010
(B) 1011
13
14
(C) 10
(D) 10
90. A body dropped from a certain height attains the same
velocity as another falling with an initial velocity u
from a height h below the first body. If g is the acceleration due to gravity, then
A
h
85. Two particles are released from the same height at
an interval of 1 s . How long after the first particle
begins to fall will the two particles be 10 m apart.
B
u
( g = 10 ms−2 )
(A) 1.25 s
(C) 2 s
1
(A)
( t1 + t2 )
2
(B)
1
( t1 − t2 )
2
t1t2
t1 + t2
(D)
t1t 2
(C)
87. The acceleration-velocity graph of a particle moving
rectilinearly is as shown in figure. Then slope of velocity-displacement graph must be
a
88. Two particles start simultaneously from the same
point and move along two straight lines, one with uniform velocity v and other with a uniform acceleration
a . If a is the angle between the lines of motion of two
particles then the least value of relative velocity will be
at time given by
v
sin α
a
(B)
v
tan α
a
(D)
v
cos α
a
v
cot α
a
89. In PROBLEM 88, the least value of relative velocity is
(B) v cos α
(A) v sin α
(C) v tan α
(D) v cot α
04_Kinematics 1_Part 3.indd 100
(B)
u = 2 gh
(C) u = 2 gh
(D) u =
gh
2
91. Two objects move uniformly toward each other. They
get closer by 4 metre each second but when they
move uniformly in the same direction, with the same
speeds, they get 4 metre closer every 10 second . The
speeds of the two objects are
(A) 2.2 ms −1 and 1.8 ms −1
(B) 1.1 ms −1 and 1.8 ms −1
(C) 22 ms −1 and 18 ms −1
92. A body falls freely under gravity. The distance travelled by it in the last second of its journey equals the
distance travelled by it in the first three second of its
free fall. The total time taken by the body to reach the
ground is
(A) 5 s
(B) 8 s
(C) 12 s
(D) 15 s
(A) increasing linearly
(B) decreasing linearly
(C) a constant
(D) increasing parabolically
(C)
(A) u = gh
(D) 1.7 ms −1 and 2.1 ms −1
v
(A)
P
(B) 1.5 s
(D) 2.5 s
86. A particle thrown down from the top of a tower takes
time t1 to reach the ground. It takes time t2 if thrown
from the same point with the same speed in the
upward direction. The time taken by it to fall freely to
the ground from the top of tower is
x
93. Six persons are initially at the six corners of a hexagon of side a . Each person now moves with a uniform
speed v in such a manner that 1 is always directed
towards 2, 2 towards 3, 3 towards 4 and so on. The
time after which they meet is
(A)
2a
v
(B)
a
v
(C)
2a
3v
(D)
a
2v
94. Velocity versus displacement graph of a particle moving in a straight line is as shown in figure. The acceleration of the particle is
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Chapter 4: Kinematics I
99. A body moving rectilinearly traversed one third of
the total distance with a velocity 4 ms −1 . The remaining part of the distance was covered with a velocity
2 ms −1 for half the time and with velocity 6 ms −1 for
the other half of time. The mean velocity averaged
over the whole time of motion is
v
x
(A) constant
(B) increases parabolically with x
(C) increases linearly with x 2
(D) increases linearly with x
95. The acceleration time graph of a particle moving in a
straight line is as shown in figure. The velocity of the
particle at time t = 0 is 2 ms −1 . The velocity after 2
second will be
(A) 5 ms −1
(B)
(C) 3.5 ms −1
(D) 4 ms −1
(C)
4
3
t(s)
(A) 2 ms −1
(B)
4 ms −1
(C) 6 ms −1
(D) 8 ms −1
96. A ball is thrown vertically up with a speed of 20 ms −1.
It is caught on its way down 5 m above the point
from where it was thrown. The time lapse between the
throw and the catch is
(A) 1.9 s
(B)
(C) 8.3 s
(D) 1.2 s
3.8 s
97. Two particles start moving from the same point along
the same straight line. The first moves with constant
velocity 2v and the second with constant acceleration a . During the time that elapses before the second
catches the first, the greatest distance between the particles is
(A)
v2
a
(B)
v2
2a
(C)
2v 2
a
(D)
v2
4a
4.5 ms −1
100. A, B, C and D are four collinear points such that
AB = BC = CD . If the average value of velocities
between A and B, C and D are 12 ms −1 and 20 ms −1
respectively, then the value of average velocity
between B and C if the body moves with uniform
acceleration throughout is
(A) 16 ms −1
a(ms–2)
1
4.101
(B) 14 ms −1
( 1 + 241 ) ms −1
(D) 8 ms −1
101. A target is made of two plates, one of wood and the
other of iron. The thickness of the wooden plate is 4
cm and that of iron plate is 2 cm. A bullet fired goes
through the wood first and then penetrates 1 cm into
iron. A similar bullet fired with the same velocity
from opposite direction goes through iron first and
then penetrates 2 cm into wood. If a1 and a2 be the
retardations offered to the bullet by wood and iron
plates respectively then
(B) a2 = 2 a1
(A) a1 = 2 a2
(C) a1 = a2
(D) Data Insufficient
102. A ball is thrown from the top of a tower of height
80 metre with a horizontal velocity of 30 ms −1 . The
velocity with which it strikes the level ground is
(A) 20 ms −1
(B)
50 ms −1
(C) 80 ms −1
(D) 100 ms −1
103. A ball is dropped vertically from a height d above
the ground. It hits the ground and bounces up verti1
cally to a height d . Neglecting subsequent motion
2
and air resistance, its velocity v varies with the
height h above the ground as
(A)
v
(B)
d
98. Starting from rest a particle moves in a straight line
v
h
d
h
with acceleration a = [ 2 + t − 2 ] ms −2 . Velocity of par(C)
ticle at the end of 4 s will be
(A) 8 ms
−1
(C) 16 ms −1
04_Kinematics 1_Part 3.indd 101
(B) 12 ms
−1
v
(D)
d
h
v
d
h
(D) 20 ms −1
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4.102 JEE Advanced Physics: Mechanics – I
104. A ball is dropped from the roof of a tower of height
h . The total distance covered by it in the last second
of its motion is equal to the distance covered by it
in first three second. The value of h in meters is
g = 10 ms −2
(
(A) 80
(C) 125
The angle with the stream direction at which the boat
must move to minimise drifting is
1
(A) sin −1 ⎛⎜ ⎞⎟
⎝ η⎠
)
(B) 100
(D) 200
(C)
105. A stone falls freely from a point O . It passes through
the points P, Q , R,..........such that OP , OQ , OR ,
.......... are in geometric progression. Then velocities of
stone at P , Q , R , ..........are in
(A) arithmetic progression
(B) geometric progression
(C) harmonic progression
(D) logarithmic mean
(A) 4 h = 3 H
(B)
(D) 4h = H
⎛ 1⎞
cot −1 ⎜ ⎟
⎝ η⎠
(D)
π
⎛ 1⎞
+ cot −1 ⎜ ⎟
⎝ η⎠
2
108. Sachin ( S ) hits a ball along the ground with a speed
u in a direction which makes an angle 30° with the
line joining him and the fielder Prem ( P ) . Prem runs
2u
. At what angle
3
θ should he run to intercept the ball?
to intercept the ball with a speed
S
106. A particle projected vertically upwards attains a maximum height H . If the ratio of the times to attain a
1
then
height h ( h < H ) is
3
(C) 3h = H
π
⎛ 1⎞
+ sin −1 ⎜ ⎟
⎝ η⎠
2
(B)
30°
θ
u
3h = 4H
107. The velocity of a boat in still water is h times
less than the velocity of flow of the river ( η > 1 ) .
P
2u
3
⎛ 3⎞
(A) sin −1 ⎜
⎝ 2 ⎟⎠
(B)
⎛ 3⎞
(C) sin −1 ⎜ ⎟
⎝ 4⎠
⎛ 4⎞
(D) sin −1 ⎜ ⎟
⎝ 5⎠
⎛ 2⎞
sin −1 ⎜ ⎟
⎝ 3⎠
Multiple Correct Choice Type Questions
This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE is/are correct.
u+v
(A)
1. A particle moves with an initial velocity v0 and
(B) u + at
2
retardation α v , where v is its velocity at any time
t . If log e ( 2 ) = 0.7 , then which of the following
statement(s) is/are correct?
(A) The particle will cover a total distance
v0
.
α
(C)
3.
1
.
α
(C)The particle will continue to move for a very long
time.
v
(D)The velocity of the particle will become 0 after
2
7
.
time
10α
Average velocity of a particle moving in a straight line,
with constant acceleration a and initial velocity u
and final velocity v in first t second is
04_Kinematics 1_Part 3.indd 102
(D) u +
1
at
2
The velocity of a particle moving along a straight
line increases according to the linear law v = v0 + kx ,
where k is a constant. Then
(A) the acceleration of the particle is k ( v0 + kx ) .
(B) The particle will come to rest after time
2.
1
( u + at )
2
(B)the particle takes a time
velocity v1 .
1
⎛v ⎞
log e ⎜ 1 ⎟ to attain a
k
⎝ v0 ⎠
(C)velocity varies linearly with displacement with
slope of velocity displacement curve equal to k.
(D) data is insufficient to arrive at a conclusion.
4.
A rod of length l leans by its upper end against a
smooth vertical wall, while its other end leans against
the floor. The end that leans against the wall moves
uniformly downward . Then the
11/28/2019 7:11:12 PM
Chapter 4: Kinematics I
4.103
average velocity in this case and s1 the total displacement. Now, the same particle, starting again from rest
is accelerated for the same time t1 with constant acceleration 2 a1 and finally comes to rest with constant
retardation a2 in time t3 . If v2 is the average velocity
in this case and s2 the total displacement. Then
vy = v
y
vy
l
vx
O
x
(A)other end moves uniformly forward with speed
v.
(B)
other end moves with a speed whose value
decreases with increase in y and vanishes at
y=0.
8.
(C)
other end moves with a speed whose value
decreases with decrease in y and vanishes at
y=0.
v
(D)other end moves such that the ratio x equals
v
y
.
l2 − y 2
5.
6.
(B)
2s1 < s2 < 4 s1
(C) v2 = 2v1
(D) 2v1 < v2 < 4v1
A particle moves with an initial velocity v0 and retardation α v , where v is velocity at any instant t . Then
v
(A) the particle will cover a total distance 0 .
α
(B)the particle continues to move for a long time
span.
1
1
(C) the particle attains a velocity v0 at t = .
α
2
(D) the particle comes to rest at t =
1
.
α
For a particle moving in a plane, if v and a be the
instantaneous velocity and acceleration, then rate of
dv
, of the particle equal(s)
change of speed,
dt
(A) a
(B) the component of a perpendicular to v
a⋅v
(C)
v
(D) the projection of a along v
9.
The position of a particle travelling along x-axis is
given by xt = t 3 − 9t 2 + 6t where xt is in cm and t is
in second. Then
10. An aeroplane flies along a straight line from A to B
with a speed v0 and back again with the same speed
v0 . A steady wind v is blowing. If AB = l then
(A)the body comes to rest firstly at ( 3 − 7 ) s and
then at ( 3 + 7 ) s .
(B)the total displacement of the particle in travelling
from the first zero of velocity to the second zero of
velocity is zero.
(C)the total displacement of the particle in travelling
from the first zero of velocity to the second zero of
velocity is –74 cm.
(D)the particle reverses its velocity at ( 3 − 7 ) s and
then at ( 3 + 7 ) s and has a negative velocity for
(3 − 7 ) < t < (3 + 7 )
7.
(A) s2 = 2s1
A particle, starting from rest is first accelerated for
time t1 with constant acceleration a1 and then stops
in time t2 with constant retardation a2 . Let v1 be the
04_Kinematics 1_Part 3.indd 103
A particle moving along a straight line with uniform
acceleration has velocities 7 ms −1 at A and 17 ms −1
at B . C is the mid-point of AB . Then
(A)the average velocity between C and B is 15 ms −1
(B)the ratio of time to go from A to C and that from
C to B is 3 : 2
(C) the velocity at C is 10 ms −1
(D)
the average velocity between A and C is
10 ms −1
(A)total time for the trip is
along the line AB .
(B)total time for the trip is
2v0l , if wind blows
v02 − v 2
2l
v02 − v 2
, if wind blows
perpendicular to the line AB .
(C)total time for the trip decreases because of the
presence of wind.
(D)total time for the trip increases because of the
presence of wind.
11. Acceleration of a particle which is at rest at x = 0 is
a = ( 4 − 2x ) i . Select the correct alternative(s)
(A) maximum speed of particle is 4 units
(B) particle further comes to rest at x = 4
11/28/2019 7:11:20 PM
4.104 JEE Advanced Physics: Mechanics – I
(C) particle oscillates about x = 2
(D) all of the above
Speed (in m/s)
12. At the instant a motor bike starts from rest in a given
direction, a car overtakes the motor bike, both moving
in the same direction.
A
60
40
C
(A) total distance travelled by the car is 500 m .
(B)maximum speed attained during the journey is
25 ms −1 .
B
(C) car travels with uniform speed for 15 s .
D
(D)car accelerates for 5 s and decelerates also for
5s.
20
O
3 6 9 12 15 18 21 24 27
time (in s)
The speed time graphs for motor bike and car are represented by OAB and CD respectively. Then
(A)at t = 18 s the motor bike and car are 180 m
apart.
(B)at t = 18 s the motor bike and car are 720 m
apart.
(C)the relative distance between motor bike and car
reduces to zero at t = 27 s and both are 1080 m
far from origin.
(D)the relative distance between motor bike and car
always remains same.
13. A car is moving rectilinearly on a horizontal path
with acceleration a0 . A person sitting inside the car
observes that an insect S is crawling up the screen
with an acceleration a . If θ is the inclination of the
screen with the horizontal, then the acceleration of the
insect
(A) perpendicular to screen is a0 tan θ
(B) perpendicular to screen is a0 sin θ
(C) along the horizontal is a0 − a cos θ
(D) parallel to screen is a + a0 cos θ
14. A particle having a velocity v = v0 at t = 0 is decelerated at the rate a = α v , where a is a positive
constant.
2 v0
(A) The particle comes to rest at t =
.
α
(B) The particle will come to rest at infinity.
(C)The distance travelled by the particle is
32
2v0
α
32
(D)The distance travelled by the particle is
04_Kinematics 1_Part 3.indd 104
15. A car starts moving rectilinearly (initial velocity zero)
first with an acceleration of 5 ms −2 then uniformly
and finally decelerating at the same rate till it stops.
Total time of journey is 25 s and average velocity during the journey is 72 kmh −1 . Then
.
2 v0
.
3 α
16. For a moving body, which of the following statement(s)
is/are true?
(A)If its speed changes but direction of motion does
not change, its velocity may remain constant.
(B)If its speed changes, its velocity must change and
it must have some acceleration.
(C)If its velocity changes, its speed must change and
it must have some acceleration.
(D)If its velocity changes, its speed may or may not
change, and it must have some acceleration.
17. A body is moving along a straight line. Its distance xt
from a point on its path at a time t after passing that
point is given by xt = 8t 2 − 3t 3 , where xt is in metre
and t is in second.
(A)
Average speed during the interval t = 0 s to
t = 4 s is 20.21 ms −1 .
(B)Average velocity during the interval t = 0 s to
t = 4 s is −16 ms −1 .
16
(C)
The body starts from rest and at t =
s it
9
reverses its direction of motion at xt = 8.43 m
from the start.
(D) It has an acceleration of −56 ms −2 at t = 4 s .
18. Let r be the radius vector of a particle in motion about
some reference point and r its modulus. Similarly v
be the velocity vector and v its modulus, then
(A)
dr ≠ dr
(C) v =
dr
dt
dr
dt
dr
(D) v =
dt
(B)
v≠
19. Two particles P and Q move in a straight line AB
towards each other. P starts from A with velocity u1
and an acceleration a1 . Q starts from B with velocity
u2 and acceleration a2 . They pass each other at the
midpoint of AB and arrive at the other ends of AB
with equal velocities.
11/28/2019 7:11:27 PM
Chapter 4: Kinematics I
(A) They meet at midpoint at time t =
2 ( u2 − u1 )
.
( a1 − a2 )
(B)
The length of path specified i.e.
l=
4 ( u2 − u1 ) ( a1u2 − a2u1 )
( a1 − a2 )
2
AB
is
24. For a particle moving rectilinearly, the velocity
time ( v -t ) graph is plotted. Which of the following
argument(s) is/are correct to explain the facts about its
motion?
v
.
(C)
They reach the other ends of AB with equal
velocities if (u2 + u1 )( a1 − a2 ) = 8( a1u2 − a2u1 ).
(D)
They reach the other ends of AB with equal
velocities if (u2 − u1 )( a1 + a2 ) = 8( a2u1 − a1u2 ).
20. Consider a body moving rectilinearly under the influ
ence of constant acceleration a . Let v denote the
velocity of the body at any instant of time. Which of
the following argument(s) is/are correct?
(A) Speed must decrease when a is negative.
(B)Speed must increase when a is negative for the
body that starts from rest initially.
(C)Speed will increase when both v and a are both
negative.
(D)Speed will decrease when v is negative and a
positive.
21. The co-ordinate of the particle in x -y plane are given
as x = 2 + 2t + 4t 2 and y = 4t + 8t 2 . The motion of the
particle is
(A) along a parabolic path
(B) non-uniformly accelerated
(C) along a straight line
(D) uniformly accelerated
22. Two particles A and B are located in x -y plane at
points ( 0 , 0 ) and ( 0 , 4 m ) . They simultaneously start
moving with velocities v = 2j ms −1 and v = 2i ms −1.
A
B
Select the correct alternative(s)
(A) the distance between them is constant
(B)
the distance between them first decreases and
then increases
(C)time after which they are at minimum distance is
1s
(D)the shortest distance between them is 2 2 m
23. Consider a body moving rectilinearly with velocity v
under the influence of an acceleration a. Which of the
following statement(s) is/are correct?
(A)The direction of a must have some correlation
with the direction of v .
(B) a can be non-zero when v = 0
(C) a must be zero when v = 0
(D) a may be zero when v ≠ 0
04_Kinematics 1_Part 3.indd 105
4.105
T
O
t
2T
(A)The acceleration of the particle remains constant.
(B)The particle changes its direction of motion at
some point.
(C)The initial and final speeds of the particle are the
same.
(D) The displacement of the particle is zero.
25. Two stationary objects when seen by an observer that
moves with a constant speed along the line joining
them (the stationary objects) will
(A) move in the same direction
(B) move in opposite directions
(C) have the same velocity
(D) have the same speed
26. A particle moving along x-axis has its velocity ( v )
varying with x co-ordinate ( x ) as v = x . Then
(A) initial velocity of particle is zero
(B) motion is uniformly accelerated
1
(C) acceleration of particle at x = 2 m is ms −2
2
(D) acceleration of particle at x = 4 m is 1 ms −2
27. Displacement time graph of a particle moving in a
straight line is as shown in figure
s
C
D
B
A
t
(A) in region A acceleration is positive
(B) in region B acceleration is negative
(C) in region C motion is uniform
(D) in region D acceleration is negative
28. A particle moves on a straight line position at any time
t is given by x = x0 e − kt , where k is a constant. Select
the correct statement(s).
11/28/2019 7:11:32 PM
4.106 JEE Advanced Physics: Mechanics – I
(A) Distance moved is infinite.
(B) Distance moved during total motion is finite
(C) Average speed for total motion is zero.
(D) Average speed for total motion is infinite.
29. The motion of the body starting from rest is governed
dv
= − v 2 + 2v − 1 , where v is speed in
by the relation
dt
ms −1 and t is time in second, then select the correct
statement(s).
−1
(A) Terminal velocity is 1 ms .
(B) The magnitude of initial acceleration is 1 ms–2.
1
.
(C) Instantaneous speed is v = −
1+ t
(D)The speed is 1.5 ms −1 when acceleration is one
fourth of its initial value
30. In the figure is shown the position of a particle moving
on the x-axis as a function of time. Then
x
20
10
2
4
6
8
t(s)
(A) the particle has come to rest for 6 times
(B) the maximum speed is at t = 6 s
(C) the velocity remain positive for t = 0 to t = 6 s
(D)the average velocity for the total period shown is
negative
31. A particle moves with an initial velocity v0 and retardation α v , where v is its velocity at any time t . Select
the correct statement(s).
v
(A) The particle will cover a total distance 0
α
1
α
(C)The particle will continue to move for a very long
time
v
(D)The velocity of the particle will become 0 after
2
1
a time
α
v 2(m2s–2)
100
25
(A)Acceleration of the particle is 15 ms −1 at t =
04_Kinematics 1_Part 3.indd 106
1
s
2
(B)Acceleration of the particle is 7.5 ms −1 at t = 1 s
(C) Acceleration of the particle is constant
(D) At t = 1 s , velocity of particle is 12.5 ms −1
33. Mark the correct statement for a particle going on a
straight line
(A)
If the velocity and acceleration have opposite
sign, the object is slowing down.
(B)If the position and velocity have opposite sign,
the particle is moving towards the origin.
(C)If the velocity is zero at an instant, the acceleration should also be zero at that instant.
(D)
If the velocity is zero for a time interval, the
acceleration is zero at any instant within the time
interval.
34. Acceleration vs time graph for a particle moving in
straight line is as shown in figure. If particle starts
from rest at t = 0 , then which of the following curve
is true for the same particle
a(ms–2)
1
1
(A)
v
(B) The particle comes to rest after a time
32. A particle starts moving rectilinearly from the origin
along the x-axis. The graph between the square of
speed and position of the particle is given in the figure.
Select the correct statement(s).
x(m)
5
1s
(C)
2s
2
t(s)
(B)
v
t
s
(D)
1s
2s
t
t
1s
s
1s
2s
t
35. A train accelerates from rest for time t1 , at a constant
acceleration α for distance x . Then it decelerates to
rest at constant retardation β in time t2 for distance
y . Then
11/28/2019 7:11:36 PM
Chapter 4: Kinematics I
(A)
x
β
=
y
α
(B)
(C)
x
t
= 1
y
t2
(D) x = y
v
38. A car accelerates from rest at a constant rate of 2 ms −2
for some time. Then it retards (speed decrease) at a
constant rate of 4 ms −2 and comes to rest. It remains
in motion for a time of 6 s.
(A) Its maximum speed is 8 ms −1
(B) Its maximum speed is 6 ms −1
2
O
(C)
Distance is always greater than magnitude of
displacement
(D)
Speed is always greater than magnitude of
velocity
β
t
= 1
α
t2
36. From v -t graph shown in figure. We can draw the
­following conclusion
4.107
1
3 4 5 6
t
(A)between t = 1 is to t = 2 s speed of particle is
decreasing
(B)between t = 2 s to t = 3 s speed of particle is
increasing
(C)between t = 5 to t = 6 s acceleration of particle is
negative
(D)between t = 0 to t = 4 s particle changes, its
direction of motion twice
(C) It travelled a total distance of 24 m
(D) It travelled a total distance of 18 m
39. A particle moves along a straight line and its velocity
depends on time as v = 4t − t 2 . Then for first 5 s , the
(A) Average velocity is
25
ms −1
3
(B) Average speed is 10 ms −1
(C) Average velocity is
5
ms −1
3
(D) Acceleration is 4 ms −2 at t = 0
37. Which of the following statement(s) is/are incorrect ?
(A) Distance and speed can never be negative
(B) Distance and speed may decrease or increase
Reasoning Based Questions
This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as
Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.
Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1.
Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.
Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.
1.
Statement-1: In a uniformly accelerated motion, acceleration time graph is straight line with positive slope.
Statement-2: Acceleration is rate of change of velocity.
2.
Statement-1: A body having non-zero acceleration can
have a constant velocity.
Statement-2: Acceleration is the rate of change of
velocity.
3.
Statement-1: Irrespective of the kind of motion possessed by a body, the body will always stay at rest in a
reference frame attached to the body itself.
Statement-2: The relative velocity of a body with
respect to itself is always zero.
04_Kinematics 1_Part 3.indd 107
4.
Statement-1: The instantaneous velocity does not
depend on instantaneous position vector.
Statement-2: The instantaneous velocity and average
velocity of a particle are always same.
5.
Statement-1: A balloon ascends from the surface of
earth with constant speed. When it was at a height
50 m above the ground, a packet is dropped from it.
To an observer on the balloon, the displacement of the
packet, from the moment it is dropped to the moment
it reaches the surface of earth, is 50 m .
Statement-2: Displacement (vector) depends upon the
reference frame used to measure it.
11/28/2019 7:11:39 PM
4.108 JEE Advanced Physics: Mechanics – I
6.
Statement-1: A man who can swim at a speed v relative to water wants to cross the river of width d , flowing with speed u . He cannot reach a point P just
opposite to him across the river, if u > v .
Statement-2: The time to reach the opposite point P
across the river is
d
2
v − u2
come out to be imaginary.
and if u > v time will
7.
Statement-1: When a particle moves along a straight
line magnitude of its average velocity is equal to its
average speed over any time interval.
Statement-2: For one dimensional motion displacement and distance may or may not be equal.
Statement-1: If two particles, moving with constant
velocities are to meet, the relative velocity must be
along the line joining the two particles.
Statement-2: Relative velocity means motion of one
particle as viewed from the other.
Statement-2: Relative acceleration is zero, whereas
relative velocity non-zero in the above situation.
dv
d 10. Statement-1:
=
v , where v has its usual
dt
dt
meaning.
Statement-2: Acceleration is the rate of change of
velocity.
11. Statement-1: x − t graph, for a particle undergoing
rectilinear motion, can be as shown in the figure.
x
8.
9.
Statement-1: Two balls are dropped one after the other
from a tall tower. The distance between them increases
linearly with time (that elapses after the second ball is
dropped and before the first hits ground).
t
Statement-2: Infinitesimal changes in velocity are
physically possible only in infinitesimal time.
12. Statement-1: Area under velocity-time graph gives
displacement.
Statement-2: Area under acceleration-time graph
gives average velocity.
Linked Comprehension Type Questions
This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a
Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For
the sake of competitiveness there may be a few questions that may have more than one correct options)
Comprehension 1
Two particles A and B start from rest at the origin x = 0
and move along a straight line such that aA = ( 6t − 3 ) ms −2
and aB = ( 12t 2 − 8 ) ms −2 , where t is in seconds. Based on
the above facts, answer the following questions.
1.
Total distance travelled by A at t = 4 s is
(A) 40 m
(B) 41 m
(C) 42 m
(D) 43 m
2.
Total distance travelled by B at t = 4 s is
(A) 192 m
(B) 184 m
(C) 196 m
(D) 200 m
3.
The distance between them at t = 4 s is
(A) 144 m
(B) 148 m
(C) 152 m
(D) 156 m
Comprehension 2
A particle initially at x = 10 m , starts moving along the
positive x -axis with an initial velocity of 40 ms −1 under
04_Kinematics 1_Part 3.indd 108
the influence of an acceleration of 10 ms −2 directed along
the negative x direction. Based on this information, answer
the following questions.
4.
The particle reverses its direction of motion at time t0
from the start. Then
(A) t0 = 2 s
(B)
t0 = 4 s
(C) t0 = 6 s
(D) t0 = 8 s
5.
The maximum x -coordinate of the particle is
(A) 30 m
(B) 60 m
(C) 90 m
(D) 120 m
6.
The velocity of the particle (in ms −1 ) at the origin of
the coordinate system is
(A) 30 2 along +x direction
(B)
30 2 along −x direction
(C) 10 7 along +x direction
(D) 10 7 along −x direction
11/28/2019 7:11:43 PM
Chapter 4: Kinematics I
7.
The time at which the particle crosses the origin is
(4 − 3 2 ) s
(C) ( 4 − 7 ) s
(A)
(4 + 3 2 ) s
(D) ( 4 + 7 ) s
(B)
Comprehension 3
Starting from rest, a particle moves rectilinearly along
x-axis, under the influence of acceleration a which varies
with time t (in second) as a = ( 2t − 4 ) ms . Based on this
information answer the following questions.
−2
8.
The particle comes to rest at time equal to
(A) 1 s
(B) 2 s
(C) 3 s
(D) 4 s
9.
The maximum velocity of the particle is vmax at time
t0 (say). Then
(A) vmax = 2 ms −1 , along +x axis at time t0 = 2 s
(B)
vmax = 4 ms −1 , along −x axis at time t0 = 2 s
(C) vmax = 2 ms −1 , along −x axis at time t0 = 4 s
(D) vmax = 4 ms −1 , along +x axis at time t0 = 4 s
10. The velocity-time graph of the particle is a
(A) Straight line passing through ( 0 , 0 )
(B) Straight line passing through ( 2, 0 )
(C) Parabola having origin at ( 4 , 2 )
(D) Parabola having origin at ( 2, 4 )
14. The acceleration ( a ) of the body as the function of
time is
(A) a = 2e −3t
(B)
The motion of a body falling initially from rest in a resistive
medium is described by the differential equation
dv
= 6 − 3v
dt
where v is the velocity of the body at any instant ( in ms −1 ).
Based on the above facts, answer the following questions.
(C) a = 6 e −3t
(D) a = 6 e −2t
Comprehension 5
A particle moves in the xy plane, such that at any instant t,
its x and y coordinates are given by
x = a sin ( ωt )
and
y = a [ 1 − cos ( ωt ) ]
where a and ω are constants. Based on the above facts,
answer the following questions.
15. The magnitude of the velocity of the particle is
(A) aω
(B)
aω sin ( ωt )
(C) aω cos ( ωt )
(D) aω [ sin(ωt) + cos(ωt) ]
16. The equation of trajectory followed by the particle is
(A) x = 2 ay
(B)
(C) x = 4 ay 2
(D) None of these
(C) 2 ms −2
(D) 18 ms −2
(B)
aω 2
aω 2
2
(D)
aω 2
3
(C)
Comprehension 6
A small ball is pushed with a speed v from A . It moves on
a smooth surface and collides with the wall at B at distance
d from A . During impact loses one third of its velocity.
3 ms −2
(A) 6 ms −1
(B)
(C) 2 ms −1
(D) 18 ms −1
v
3 ms −1
1
(C) v = ( 1 − e −3t )
2
04_Kinematics 1_Part 3.indd 109
(B)
v = 1− e
A
B
Based on the above facts, answer the following questions.
18. The average velocity of the ball during its motion from
A to B and back to A will be
13. The velocity at any time t is given by
(A) v = e
y
2a
d
12. The terminal velocity i.e., the velocity at which acceleration becomes zero is given by
−3 t
x = 2 ay 1 −
(A) aω
11. The initial acceleration is of the body is
(B)
a = 3 e −2t
17. The magnitude of the acceleration of the particle is
Comprehension 4
(A) 6 ms −2
4.109
−3 t
(D) v = 2 ( 1 − e −3t )
(A) zero
(B)
v
2
v
3
(D)
2v
3
(C)
11/28/2019 7:11:52 PM
4.110 JEE Advanced Physics: Mechanics – I
19. The total time taken by the ball in moving from A to B
and back to A is T . Then T equals
(A)
2d
v
(B)
3d
v
(C)
4d
v
5d
(D)
2v
20. The average speed during the journey from A to B
and back to A is
v
2v
(B)
(A)
5
5
(C)
3v
5
4v
(D)
5
Comprehension 7
A car accelerates from rest with 2 ms −2 on a straight track
and then it comes to rest applying its brakes. The total distance travelled by the car is 100 m in 20 s. Based on the
above facts, answer the following questions.
21. The maximum speed attained by the car is
(A) 5 ms −1
(B) 10 ms −1
(C) 15 ms −1
(D) 20 ms −1
22. The duration for which the brakes were applied is
(A) 10 s
(B) 5 s
(C) 15 s
(D) 16 s
23. The maximum retardation given to the car is
(A) 2 ms −2
(C)
4
ms −2
3
2
ms −2
3
5
ms −2
(D)
3
(B)
24. The average speed of the car for the entire tenure of
motion is
(A) 5 ms
−1
(B)
(C) 8 ms
−1
(D) 8.2 ms −1
6 ms
−1
25. The distance covered during acceleration is
(A) 25 m
(B) 15 m
(C) 10 m
(D) 5 m
26. The distance covered during retardation is
(A) 85 m
(B) 75 m
(C) 50 m
(D) 25 m
Comprehension 8
A body falling from a height H hits an inclined plane in
its path at a height h ( < H ) . As a result of this impact, the
direction of the velocity of the body becomes horizontal.
Based on the above facts, answer the following questions.
04_Kinematics 1_Part 3.indd 110
h
for which the body will take the maxiH
mum time to reach the ground is
27. The value of
(A)
1
2
(B)
(C)
1
2
(D) 2
2
28. The time taken by the body to hit the ground is
(A) T = 2
H
g
(B)
(C) T =
2H
g
(D) T =
T=4
H
g
3H
g
Comprehension 9
A body is projected from the ground vertically upwards.
The body is observed to be at height h above the ground
at two times t1 and t2 while ascending and descending
respectively. Based on the above facts, answer the following questions.
29. The height h in terms of t1 and t2 is
(A) h = gt1t2
1
gt1t2
2
(C) h =
(B)
h = 2 gt1t2
(D) h =
1
gt1t2
4
30. The velocity of projection ( u ) must be
1
g ( t1 + t2 )
2
(A) u =
(C) u = 2 g ( t1 + t2 )
(B)
u=
1
g ( t1 + t2 )
4
(D) u = 4 g ( t1 + t2 )
31. The maximum height ( H ) reached by the body is
(A) H =
1 2
gt1
2
(B)
(C) H =
1
2
g ( t1 + t2 )
8
(D) H =
H=
1 2
gt2
2
1
2
g ( t1 + t2 )
4
32. The velocity ( v ) of the body at height
(A) v =
1
gt1
4
(B)
(C) v =
1
g t12 + t22
2
(D) v =
v=
h
is
2
1
gt2
4
1
g t1t2
4
11/28/2019 7:11:58 PM
Chapter 4: Kinematics I
33. The velocity of the particle at half the maximum
height.
(A)
(C)
g 2 2
t1 + t2
2
g
2 2
( t1 + t2 )
(
)
(B)
g 2 2
t1 + t2
4
(D)
g
( t1 + t2 )
2
Comprehension 10
v (in ms–1)
If the velocity v of a particle moving along a straight line
decreases linearly with its displacement s from 20 ms −1 to
a value approaching zero at s = 30 m. Based on the above
facts, answer the following questions.
20
38. How many buses the bus A will meet on way from city
A to city B ?
(A) 6
(B) 5
(C) 12
(D) 11
Comprehension 12
A particle moves in positive x-direction according to
law x = 12t − t 2 m . where t time in second. (Take +ve xdirection as +ve).
Based on the above facts, answer the following
questions.
39. Average velocity from t = 0 to t = 8 s is
(A) 4 ms −1
(B)
6 ms −1
(C) 8 ms −1
(D) −4 ms −1
40. Average speed from t = 0 to t = 8 s is
O
s (in m)
30
34. Acceleration of the particle at s = 15 m is
(A) 4 ms −1
(B)
5 ms −1
(C) 6 ms −1
(D) 8 ms −1
41. Average acceleration from t = 0 s to t = 8 s is
(A)
2
ms −2
3
(B)
−
2
ms −2
3
(A) −
1
ms −2
4
(B)
(C)
20
ms −2
3
(D) −
20
ms −2
3
(C) −
1
ms −2
2
(D) −2 ms −2
35. The time taken by the particle to reach the 30 m position is
(A) 1.5 s
(B) 3 s
(C) 6 s
(D) Infinite
Comprehension 11
Buses are going from one city to other and vice-versa. The
buses start at regular interval of 5 minutes each from station A to B and station B to A each with same speed
60 kmh −1 . The distance between two stations is 30 km.
A bus marked A starts from city A and finds buses
approaching from opposite direction.
Based on the above facts, answer the following
questions.
36. The time interval after which bus A will meet two
buses coming from opposite side is
(A) 5 min
(B) 2.5 min
(C) 1.25 min
(D) 10 min
37. The distance travelled by bus A to meet two buses
from opposite side is
(A) 2.5 km
(B) 5 km
(C) 10 km
(D) None of these
04_Kinematics 1_Part 3.indd 111
4.111
+
1
ms −2
4
Comprehension 13
A particle starts from rest from the origin with a time varying acceleration a = ( 2t − 4 ) , where t is in seconds and a
in ms −2 . Assuming the particle to move rectilinearly.
Based on the above facts, answer the following
questions.
42. Particle comes to rest (after a time) at
(A) 1 s
(B) 4 s
(C) 3 s
(D) 2 s
43. The speed of the particle moving in negative direction
is maximum at time t . Then t equals
(A) 3 s
(B) 4 s
(C) 2 s
(D) 1 s
44. The distance travelled by the particle from the start to
the moment when it comes to rest is
16
m
3
(C) 3 m
(A)
(B) 2 m
(D) None of these
Comprehension 14
A person standing on the roof of a building throws a ball
vertically upward at an instant t = 0 . The ball leaves his
11/28/2019 7:12:04 PM
4.112 JEE Advanced Physics: Mechanics – I
hand with an upward speed 20 ms −1 and it is then in free
fall. The ball rises to a certain height and then moves down.
On its way down, the ball just misses to hit the roof of the
building and keeps falling towards the earth. The ball hits
earth at t = 5 s . Considering that
(i) the vertically upward direction is the positive
y-direction
(ii) the position of ball at t = 0 is the origin
(iii) the ball does not rebound and comes to rest at the
same place where it hits earth and
(iv) air resistance is negligible, answer these questions.
Take g = 10 ms −2
(
)
Based on the above facts, answer the following questions.
45. Position-time graph for the given motion of the ball is
(A) y(m)
O
(B) y(m)
1 2 3 4 5
t(s)
(C) y(m)
O
O
1 2 3 4 5
t(s)
(D) y(m)
5
1 2 3 4
t(s)
O
3
1 2
4 5
t(s)
46. Velocity of the ball will vary with time as
(A) v(ms–1)
O
1 2 3
5
t(s)
(C) v(ms–1)
O
1 2 3 4 5
O
4 5
1 2 3
t(s)
(D) v(ms–1)
t(s)
O
2
1
3 4 5
t(s)
O
04_Kinematics 1_Part 3.indd 112
5
2
1
3 4
(B) a(ms–2)
t(s)
O
5
1 2 3 4
t(s)
O
2
1
3 4 5
t(s)
Comprehension 15
The position of a particle is moving along the x-axis
depends on the time according to the equation x = 6t 2 − t 3 ,
where x is in metre and t in seconds.
Based on the above facts, answer the following
questions.
48. Time at which velocity of the particle is maximum
along positive direction of x-axis is
(A) 1 s
(B) 2 s
(C) 3 s
(D) 4 s
49. Distance travelled by the particle during time interval
t = 3 s to t = 5 s is
(A) 2 m
(B) 5 m
(C) 12 m
(D) 10 m
50. Average speed of the particle during time interval
t = 0 s to t = 6 s is
(A) 2 ms −1
(B) Zero
(C) 4 ms −1
(D) None of these
2
1
3 4 5
A particle is moving along x-axis and its initial velocity is 27 ms −1 . The acceleration of particle is given by
the relation a = ( −6t ) ms −2 , where t is in seconds. At
t = 0 particle is at x = 0 .
Based on the above facts, answer the following
questions.
51. The velocity of particle, when it travels 26 m is
(B) 15 ms −1
(A) 21 ms −1
−1
(C) 24 ms
(D) 18 ms −1
52. Maximum value of velocity along positive x-direction
is
(B) 33 ms −1
(A) 35 ms −1
−1
(C) 27 ms
(D) 30 ms −1
47. Acceleration of the ball will vary with time as
(A) a(ms–2)
O
(D) a(ms–2)
Comprehension 16
(B) v(ms–1)
4
(C) a(ms–2)
t(s)
53. Maximum value of displacement along positive
x-direction is
(A) 54 m
(B) 27 m
(C) 120 m
(D) None of these
11/28/2019 7:12:08 PM
Chapter 4: Kinematics I
4.113
Matrix Match/Column Match Type Questions
Each question in this section contains statements given in two columns, which have to be matched. The statements in
COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of
bubbles will look like the following:
A
B
C
D
1.
p
p
p
p
p
q
q
q
q
q
r
s
t
r
r
r
r
s
s
s
s
t
t
t
t
From the v -t graph shown in figure, match the quantities in COLUMN-I to their respective conclusions in
COLUMN-II.
v(in ms–1)
1
0
COLUMN-II
(C) Acceleration, in ms −2 , at
t=4s
(r) Zero
(D) Speed, in ms −1 , at t = 4 s
(s) −4
(t) None of these
2
3
4
5
6
t(in s)
3.
COLUMN-I
COLUMN-II
(A) between t = 0 and t = 1 s
(p) v = 0
(B) between t = 1 s and t = 2 s
(q) a = 0
(C) between t = 2 s and t = 3 s
(r) v ≠ 0
(D) between t = 3 s and t = 4 s
(s) a ≠ 0
(E) between t = 4 s and t = 5 s
(t) accelerated
(F) between t = 5 s and t = 6 s
(u) decelerated
Match the quantities in COLUMN-I with the corresponding expressions in COLUMN-II.
COLUMN-I
COLUMN-II
(p)
dv
dt
(q)
dr
dt
(r)
dv
dt
(s)
d2r
dt 2
For a particle moving rectilinearly, the x varies with
(t)
t as per the equation x = −5t 2 + 20t + 10 , where x is
in metre and t is in second.
dv
dt
(u) None of these
(A) Velocity
(B) Tangential acceleration
(C) Acceleration
(D) Instantaneous speed
(G) at t = 1 s and at t = 3 s
2.
COLUMN-I
COLUMN-I
COLUMN-II
(A) Average speed, in ms −1 ,
from t = 0 to t = 4 s
(p) 20
(B) Average velocity, in ms −1 ,
from t = 0 to t = 4 s
(q) 10
4.
A particle moves such that its x coordinate is related
to the time t by the relation t = x + 3 , where x is
in metre, t is in second. Based on this information,
match the values in COLUMN-I (in SI units) to their
respective quantities for the particles motion given in
COLUMN-II.
(Continued)
04_Kinematics 1_Part 4.indd 113
11/28/2019 7:08:44 PM
4.114 JEE Advanced Physics: Mechanics – I
COLUMN-I
COLUMN-II
COLUMN-I
COLUMN-II
(A) 0
(p) Acceleration at t = 5 s .
(A) vinst = vav
(B) 2
(p) for uniform motion in any
direction
(q) Average speed from t = 0 to
t = 6 s.
(B) vinst = v
(q) for uniform motion in given
direction
(C) 3
(r) Velocity at the point of reversal of
motion.
(C) vinst = vav
(r) Always true
(D) 18
(s) Total distance travelled from
t = 0 to t = 6 s .
(D) vinst < v
(s) Never true
(t) Displacement from t = 0 to
t = 6 s.
A man can row a boat with 4 kmh −1 in still water. The
man wishes to cross the river of width 4 km having
a water current of 2 kmhr −1 . To cross the river with
zero drift he swims making at an angle α degree with
the current flow taking a time t1 minutes to cross the
river.
Now he wishes to cross the river in the shortest time
t2 minutes making an angle β degree with the river
flow.
Further he takes a time t3 minutes to row 2 km
upstream and then downstream back to the start
point. Assuming all the cases to be independent of
each other, the man to start from the river bank from
the same point in the first two cases and from the midpoint of the river in the third case, match the quantities
in COLUMN-I to the values in COLUMN-II.
7.
5.
COLUMN-I
COLUMN-I
COLUMN-II
(A) Motion of
dropped ball
(p) Two dimensional motion
(B) Motion of a
snake
(q) Three dimensional
motion
(C) Motion of a bird (r) One dimensional motion
(D) Earth
8.
(s) Absolute rest
The displacement-time graph of a body moving on a
straight line is given by
x
Parabola
COLUMN-II
0
T
2T
t
(A) α
(p) 40 3
(B) β
(q) 60
COLUMN-I
COLUMN-II
(C) t1
(r) 80
(A) Velocity – time graph
(p)
(D) t2
(s) 90
(E) t3
(t) 120
(u) Zero
6.
Match the following
For one dimensional motion if vav be the average
speed, vav be the average velocity, vinst be the instan
taneous speed, vinst be the instantaneous velocity and
v be the speed, then match the following
04_Kinematics 1_Part 4.indd 114
T 2T
(B) Acceleration-time graph
(q)
T
2T
(Continued)
11/28/2019 7:08:49 PM
Chapter 4: Kinematics I
COLUMN-I
COLUMN-II
(C) Distance – time graph
(r)
(D) Speed – time graph
9.
and come at rest. If all quantities are in SI units, then
match the following columns.
COLUMN-I
2T
(p)
t1
t2
(B)
α
β
(q)
t2
t1
(D) Maximum speed
attained in it whole
journey
v(ms–1)
10
2
4
6
t(s)
COLUMN-I
COLUMN-II
(A) Change in velocity
5
(p) − SI unit
3
(B) Average acceleration
(q) −20 SI unit
(C) Total displacement
(r) −10 SI unit
(D) Acceleration at t = 3 s
(s) −5 SI unit
10. A balloon rises up with constant net acceleration of 10 ms −2 . After 2 s a particle drops from
the balloon. After further 2 s match the following
Take g = 10 ms −2
)
COLUMN-I
COLUMN-II
(A) Height of particle ground
(p) Zero
(B) Speed of particle
(q) 10 SI units
(C) Displacement of particle
(r) 40 SI units
(D) Acceleration of particle
(s) 20 SI units
11. A body accelerates from rest for time t1 at a constant
rate α for distance x then it decelerates at constant
rate β for time t2 and covers distance y in this time
04_Kinematics 1_Part 4.indd 115
x
y
(C) average speed for whole
(r)
journey
For the velocity-time graph shown in figure, in a time
interval from t = 0 to t = 6 s , match the following
(
COLUMN-II
(A)
(s)
T
4.115
(s)
2αβ
(x + y)
α+β
αβ ⎛ x + y ⎞
⎜
⎟
α+β⎝ 2 ⎠
12. The equation of one dimensional motion of particle is
described in COLUMN-I. At t = 0 , particle is at origin
and at rest. Match the COLUMN-I with the statements
in COLUMN-II.
13.
COLUMN-I
COLUMN-II
(A) x = ( 3t 2 + 2 ) m
(p) velocity of particle at
t = 1 s is 8 ms −1
(B) v = 8t ms −1
(q) particle moves with
uniform acceleration
(C) a = 16t
(r) particle moves with
variable acceleration
(D) v = 6t − 3t 2
(s) particle will change its
direction some time
v -t graph of a particle moving along positive direction
x is shown in figure. Match the items in COLUMN-I
with the respective answers in COLUMN-II.
v
t
COLUMN-I
COLUMN-II
(A) a-x graph
(p) Parabola
(Continued)
11/28/2019 7:08:54 PM
4.116 JEE Advanced Physics: Mechanics – I
COLUMN-I
COLUMN-II
COLUMN-I
COLUMN-II
(B) v -x graph
(q) Circle
(A) M
(p) A–1
(C) a-t graph
(r) Straight line
(B) N
(q) R–1
(D) a-v graph
(s) Ellipse
(C) P
(r) A
(D) Q
(s) R
14. Match the v -t graphs in COLUMN-I with the respective a-t graphs in COLUMN-II.
COLUMN-I
COLUMN-II
(A) v
(p) a
t
t
–a0
(B) v
(q) a
t
t
(C) v
(r) a
t
16. In the s-t equation
following
COLUMN-I
COLUMN-II
(A) Distance travelled in 3 s
(p) –20 unit
(B) Displacement in 1 s
(q) 15 unit
(C) Initial acceleration
(r) 25 unit
(D) Velocity at 4 s
(s) –10 unit
17. The velocity time graphs for a particle moving along a
straight line is given in each situation of COLUMN–I.
Match the graph in COLUMN–I with corresponding
statements in COLUMN-II.
COLUMN-I
COLUMN-II
(A) v
(p) Speed of particle is
continuously decreasing.
t
(D) v
t
(s) a
(B) v
t
t
15. Let us call a motion, A when velocity is positive and
increasing. A −1 when velocity is negative and increasing. R when velocity is positive and decreasing and
R −1 when velocity is negative and decreasing. Now
match the following two tables for the given s-t
graph
s
t
M
04_Kinematics 1_Part 4.indd 116
(q) Magnitude of acceleration
of particle is decreasing
with time.
(r) Direction of acceleration of
particle does not change.
(C) v
t
(D) v
t
N
( s = 10 + 20t − 5t 2 ) match the
(s) Magnitude of acceleration
of particle does not change.
P
Q
t
(t) Acceleration is always
opposite to the direction of
velocity.
11/28/2019 7:08:57 PM
Chapter 4: Kinematics I
18. Match the statements in COLUMN–I with corresponding graphs in COLUMN-II.
4.117
be ­elastic, match the statements in COLUMN–I with
corresponding graphs in COLUMN-II.
COLUMN-I
COLUMN-II
COLUMN-I
COLUMN-II
(A) Particle moving with
constant speed.
(p) s
(A) The distance travelled
by particle varies with
time as
(p)
(B) Velocity of particle
changes with time as
(q)
(C) Displacement of
particle depends on
time as
(r)
(D) Dependency of
acceleration on time is
given by
(s)
t
(B) Particle moving with
increasing acceleration.
(q) s
t
(C) Particle moving with
constant negative
acceleration.
(r) v
t
(D) Particle moving with
zero acceleration.
(s) s
t
(A) Distance travelled
Dd
(p) Slope of a distancetime graph
(B) Velocity change Dv
(q) Slope of velocitytime graph
(C) Velocity v
(r) Area under a
velocity-time graph
(D) Acceleration a
(s) Area under an
acceleration-time
graph.
20. A particle is dropped vertically downward under
gravity. Consider the downward direction as positive and the collision of the ball with the ground to
04_Kinematics 1_Part 4.indd 117
COLUMN-II
(A)
(p) Variable velocity
(x)
Position
COLUMN-II
COLUMN-I
Time
(B)
(t)
(q) Positive acceleration
(v)
Velocity
COLUMN-I
21. A particle is moving along x-direction in four ways.
Different graphs is plotted in COLUMN-I.
Position
(C)
(x)
(r) Negative acceleration
(x)
Acceleration
19. The motion of an object over time can often be communicated by graphs of its distance, velocity or acceleration with time. Different features of these graphs correspond to quantities of the motion. Match each quantity
in the COLUMN-I with its graphical manifestation in
the COLUMN-II.
Time
(t)
(Continued)
11/28/2019 7:08:59 PM
4.118 JEE Advanced Physics: Mechanics – I
COLUMN-I
COLUMN-II
(D)
(s) Constant speed
Position
(x)
s
Time
(t)
t1
22. Match the following
t2
t3
t4
COLUMN-I
COLUMN-II
COLUMN-I
COLUMN-II
(A) t1
(p) a > 0
(A) Constant positive
acceleration
(p) Speed may increase
(B) t2
(q) a < 0
(B) Constant negative
acceleration
(q) Speed may decrease
(C) t3
(r) v > 0
(D) t4
(s) v < 0
(C) Constant
displacement
(r) Speed is zero
(D) Constant slope of
a-t graph
(s) Speed must increase
23. Two ships A and B are 10 km apart on a line running from south to north. A is towards north of B
and moving west with a speed of 20 kmh −1 while B
is moving towards north with 20 kmh −1 . The distance
of their closest approach in metres is l and the time in
second taken to reach this position is t .
COLUMN-I
COLUMN-II
(A) l
(p) North-West
(B) t
(q) 7071
(C) VAB
(r) 900
(D) VBA
(s) South-East
24. A particle is moving in straight line and its displacement
versus time graph is as shown in figure. COLUMN-I
contains different instant and COLUMN-II contains
values of acceleration and velocities at those instants.
Match them
04_Kinematics 1_Part 4.indd 118
t
(t) None
25. A particle is moving along x-axis. Its x-coordinate is
varying with time as x = −20t + 5t 2 . For the given
equation, match the following columns.
COLUMN-I
COLUMN-II
(A) At what time particle
changes its direction
of motion
(p) 1 s
(B) At what time
magnitude of velocity
and acceleration are
equal
(q) 2 s
(C) In how much time,
distance travelled by
the particle becomes
25 m
(r) 3 s
(D) In how much time the
displacement of the
particle becomes 15 m
(s) 4 s
(t) None of the above
11/28/2019 7:09:02 PM
Chapter 4: Kinematics I
4.119
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical value obtained after doing series of calculations based
on the data given in the question(s).
1.
If the body covers equal displacements in successive intervals of time t1 , t2 and t3 then show that
7.
A bus starts from rest with a constant acceleration of
5 ms −2 . At the same time a car travelling with a constant velocity of 50 ms −1 overtakes and passes the
bus. Find
(a)at what distance, in metre, will the bus overtake
the car?
(b)how fast, in ms −1 , will the bus be travelling then?
8.
A truck travelling along a straight road at a constant
speed of 72 kmh −1 passes a car at time t = 0 moving much slower. At the instant the truck passes the
car, the car starts accelerating at constant 1 msec −2
and overtake the truck 0.6 km further down the road,
from where the car moves uniformly. Find the distance
between them, in metre, at time t = 50 s from the start.
9.
A train travelling at 72 kmh −1 is checked by track
repairs. It retards uniformly for 200 m, covering the
next 400 m at constant speed and accelerates uniformly to 72 kmh −1 in a further 600 m . If the time
at the constant lower speed is equal to the sum of the
times taken in retarding and accelerating, find the total
time, in minutes, taken.
k
1 1 1
. Find k .
− + =
t1 t2 t3 t1 + t2 + t3
2.
Between two stations a train accelerates uniformly at
first, then moves with a constant speed and finally
retards uniformly. If the ratios of the time taken are
1 : 8 : 1 and the greatest speed attained by the train is
60 kmh −1 , find the average speed, in ms −1 , over the
whole journey.
3.
A sphere is fired downwards into a medium with an
initial speed of 27 ms −1 . If it experiences a deceleration of a = ( −6t ) ms −2 , where t is in ­seconds, determine the distance, in metre, travelled before it stops.
4.
A particle travels along a straight line such that in 2 s
it moves from an initial position x A = +0.5 m to a position xB = −1.5 m . Then in another 4 s it moves from
xB to xC = +2.5 m . Determine the particle’s average
speed, in ms −1 , during the 6 s time interval.
5.
6.
A particle moving with constant acceleration along a
straight line covers the distance between two points
80 m apart in 10 s . Its speed as at passes second
point is 18 ms −1 .
(a) What is its speed, in ms −1 , at the first point?
(b) What is its acceleration in ms −2 ?
(c)At what prior distance, in m and time, in s from
first point, the particle reverses its direction of
motion?
(d)What is the total distance travelled, in m, during
this 10 s ?
Tests reveal that a normal driver takes about 0.75 s
before he or she can react to a situation to avoid a collision. It takes about 3 s for a driver having 0.1%
alcohol in his system to do the same. If such drivers
are travelling on a straight road at 44 ms −1 and their
cars can decelerate at 2 ms −2 , determine the shortest
­stopping distance ( d ) for each, in metre, from the
moment they see the pedestrians.
11. A bullet fired into a fixed target looses half of its velocity after penetrating 3 cm . How much further, in cm ,
it will penetrate before coming to rest assuming that it
faces constant resistance to motion.
12. A body travels 200 cm in the first two second and
220 cm in the next four second. What will be the
velocity, in cms −1 at the end of seventh second from
the start?
13. A particle moves with uniform acceleration a . If v1 ,
v2 and v3 be the average velocities in three successive
intervals of time t1 , t2 and t3 respectively, then find
the value of
v1 = 44 ms–1
d
04_Kinematics 1_Part 4.indd 119
10. At the instant the traffic light turns green, a car starts
with a constant acceleration of 2 ms −2 . At the same
instant a truck travelling with a constant speed of
10 ms −1 , overtakes and passes the car. How far
beyond the starting point, in m, with the car overtake
the truck? How fast, in ms −1 , will the car be travelling
at that instant?
( v1 − v2 ) ( t3 + t2 ) .
( v2 − v3 ) ( t2 + t1 )
14. A sports car travels along a straight road with an acceleration-deceleration described by the graph. If the car
starts from rest, determine the distance x0 , in m, the
car travels until it stops.
11/28/2019 7:09:10 PM
4.120 JEE Advanced Physics: Mechanics – I
applies brakes producing a constant retardation and
just manages to avoid a collision. What is the retardation of the train A , in cms −2 ? For how long is this
retardation produced?
a(ms–2)
6
x0
1000
x(m)
–4
towards north at a speed of 12 ms −1 and sinking at a
15. A balloon rises from rest on the ground with constant
g
acceleration
. A stone is dropped when the balloon
8
has risen to a height H metre. The time taken by the
stone to reach the ground is given by x
H
. Find x .
g
16. As a train accelerates uniformly it passes successive kilometre marks while travelling at velocities
of 2 ms −1 and then 10 ms −1 . Determine the train’s
velocity, in ms −1 , when it passes the next kilometre
mark and the time it takes, in s, to travel the 2 km
distance.
17. From a point A on bank of a channel with still water
a person must get to a point B on the opposite bank.
All the distances are shown in figure. The person uses
a boat to travel across the channel and then walks
along the bank to point B. The velocity of the boat
is v1 and the velocity of the walking person is v2. If
v1 = 3 3 ms −1 , α 1 = 30° and α 2 = 60°, then for what
value of v2, in ms–1, the person takes a minimum time
to go from A to B.
A
d
α2
B
rate of 2 ms −1 . The commander of submarine observes
a helicopter ascending at a rate of 5 ms −1 and heading
towards west with 4 ms −1 . Find the actual speed of
the helicopter and its speed with respect to boat, both
in ms −1 .
21. The speed-time graph of a particle moving along a
fixed direction is shown in figure. Obtain the distance
travelled by the particle, in metre and the average
speed of the particle in ms −1 between
(a) t = 0 to 10 s
(b t = 2 to 6 s.
Speed (ms–1)
A
12
O
B
10
5
Time (s)
22. At the moment t = 0, a particle leaves the origin and
moves in the positive direction of the x-axis. Its velocity
t
varies with time as v = v0 ⎛⎜ 1 − ⎞⎟ where v0 = 10 cms −1
⎝
5⎠
is the initial speed of the particle. The particle will be
at the distance of 10 cm from the origin at three different instants. Find out the approximate time interval
between the second and the third instant.
a
α1
b
18. A person walks up a stationary 15 m long escalator in
90 s. When standing on the same escalator, now moving, the person is carried up in 60 s. How much time,
in seconds, would it take that person to walk up the
moving escalator? Does the answer depend on the
length of the escalator?
19. The driver of the train A moving with a speed of
144 kmh −1 sights another train B 1 km ahead of
him. The train B is moving with a uniform speed of
108 kmh −1 . The driver of the train A immediately
04_Kinematics 1_Part 4.indd 120
20. A sailor in a boat, which is going due east with a
speed of 8 ms −1 observes that a submarine is heading
23. An athlete takes 2 s to reach his maximum speed of
36 kmh −1 . The magnitude of his average acceleration
(in ms −2 ) is ………
24. The diagram shows variation of
−1
time (where v is in ms ).
1
with respect to
v
1 s
v m
3
45°
3
t(s)
11/28/2019 7:09:16 PM
Chapter 4: Kinematics I
m⎞
⎛
Find the instantaneous acceleration ⎜ in 2 ⎟ of body
⎝
s ⎠
at t = 3 s .
25. A ball is thrown upwards from the foot of a tower.
The ball crosses the top of tower twice after an interval of 4 seconds and the ball reaches ground after
8 seconds , then the height of tower in meters is
( g = 10 ms −2 )
26. An insect moves with a constant velocity v from one
corner of a room to other corner which is opposite of
the first corner along the largest diagonal of room.
If the insect cannot fly and dimensions of room is
a × a × a , then the minimum time in which the insect
a
can move is
times the square root of a number n ,
v
then n is equal to?
27. A body moves with constant acceleration covers 16 m
and 24 m in successive intervals of 4 s and 2 s . Then
its acceleration in ms −2 is ………
28. Figure shows the graph of velocity versus time for a
particle going along x-axis. Initially at t = 0 , particle
is at x = 3 m . The position of the particle at t = 2 s
( in m ) is ………
4.121
32. A bullet going with speed 16 ms −1 enters a concrete
wall and penetrates a distance of 0.4 m before coming to rest. Then the time taken during the retardation
is x × 10 −2 s . Find x .
33. A baseball is moving at 25 ms −1 when it is struck by a
bat and moves off in the opposite direction at 35 ms −1.
If the impact lasted 0.010 s , find the baseball’s acceleration during the impact. ( in kms −2 )
34. A point moves with uniform acceleration and its initial
speed and final speed are 2 ms −1 and 8 ms −1 respectively then, the space average of velocity ( in ms −1 )
over the distance moved is ………
35. A man is running with a speed 8 ms −1 constant in
magnitude and direction passes under a lantern hanging at a height 10 m above the ground. Find the velocity which the edge of the shadow of the man’s head
moves over the ground with if his height is 2 m .
36. Two men P and Q are standing at corners A and B of
square ABCD of side 8 m. They start moving along the
tank with constant speed 2 ms −1 and 10 ms −1 respectively. Find the time, in second, when they will meet
for the first time.
v(ms–1)
B
10 ms–1
C
10
2 ms–1
2
O
8
29. A swimmer jumps from a bridge over a canal and
swims 1 km up stream. After that first km, he passes a
floating cork. He continues swimming for half an hour
and then turns around and swims back to the bridge.
The swimmer and the cork reach the bridge at the
same time. Assuming the swimmer had been swimming at a constant speed, calculate how fast does the
water in the canal flow in kmh −1 .
30. A car travelling at 60 kmh −1 over takes another car
travelling at 42 kmh −1 . Assuming each car to be 5 m
long. Find the time taken during the over take. (in sec)
31. A particle is moving on a straight line with constant
retardation of 1 ms −2 . What is the average speed of
the particle on the last two meters before it stops?
( in ms −1 )
04_Kinematics 1_Part 4.indd 121
D
A
t
37. A particle starting from rest undergoes acceleration given by a = t − 2 ms −2 where t is time in sec.
Velocity of particle after 4 sec is ………
38. A ball is thrown upward from the edge of a cliff with
an initial velocity of 6 ms −1 . How fast is it moving half
(
second later? g = 10 ms −2
)
39. A car goes from 20 to 30 kmh −1 in 1.5 s . At the same
acceleration, how long will it take the car to go from 30
to 36.7 kmh −1 ? (in sec)
40. The particle moves with rectilinear motion given the
acceleration-displacement ( a-S ) curve is shown in
figure. If the initial velocity is 10 ms −1 then velocity of
particle after particle has travelled 30 m divided by 5
is equal to.
11/28/2019 7:09:22 PM
4.122 JEE Advanced Physics: Mechanics – I
44. Figure shows the graph of the x-coordinate of a particle going along the x-axis as function of time.
a(ms–2)
x
10
A
8m
15
30
4m
S(m)
41. A police jeep is chasing a culprit going on a motor
bike. The motor bike crosses a turning at a speed of
72 kmh −1 . The jeep follows it a speed of 90 kmh −1
crossing the turning ten seconds later than the bike.
Assuming that they travel at constant speeds, how far
from the turning will the jeep catch up with the bike?
(in km)
42. A boy standing on a long railroad car throws a ball
straight upwards. The car is moving on the horizontal road with an acceleration of 1 ms −2 and projection
velocity in the vertical direction is 9.8 ms −1 . How far
behind the boy will the ball fall on the car? (in m)
O
4s
8s
B
t
16 s
12 s
The speed of particle at t = 12.5 s ( in ms −1 ) is ………
45. A particle moving in a straight line covers half the
distance with speed of 3 ms −1 . The other half of the
distance is covered in two equal time intervals with
a speeds of 4.5 ms −1 and 7.5 ms −1 , respectively. Find
the average speed of the particle during this motion.
46. A body initially at rest moving along x-axis in such
a way so that its acceleration displacement plot is as
shown in figure. What will be the maximum velocity
of particle in ms −1 .
a
43. The speed of a motor launch with respect to the water
is v = 5 ms −1 , the speed of stream u = 3 ms −1 . When
the launch began travelled 3.6 km up stream, turned
about and caught up with the float. How long is it
before the launch reaches the float again? (Find answer
in hour).
1 ms–2
0.5 m 1 m
s
ARCHIVE: JEE MAIN
1. [Online April 2019]
Ship A is sailing towards north-east with veloc
ity v = 30iˆ + 50 ˆj kmhr −1 where î points east and ĵ,
(I)
(II) v
a
north. Ship B is at a distance of 80 km east and
150 km north of Ship A and is sailing towards west
at 10 kmhr −1 . A will be at minimum distance B in
(A) 2.2 hr
(C) 3.2 hr
(B) 4.2 hr
(D) 2.6 hr
2. [Online April 2019]
A particle starts from origin O from rest and moves
with a uniform acceleration along the positive x-axis .
Identify all figures that correctly represent the
motion qualitatively. ( a = acceleration , v = velocity ,
x = displacement , t = time )
04_Kinematics 1_Part 4.indd 122
O
t
t
(IV) x
(III) x
O
O
t
O
t
11/28/2019 7:09:27 PM
Chapter 4: Kinematics I
(A) (I)
(C) (I), (II), (IV)
(B) (I), (II), (III)
(D) (II), (III)
3. [Online April 2019]
The stream of a river is flowing with a speed of
2 kmh −1. A swimmer can swim at a speed of 4 kmh −1.
What should be the direction of the swimmer with
respect to the flow of the river to cross the river
straight?
(B) 90°
(A) 60°
(C) 150°
(D) 120°
4. [Online April 2019]
The position vector of a particle changes with time
according to the relation r ( t ) = 15t 2iˆ + ( 4 − 20t 2 ) ˆj .
What is the magnitude of the acceleration at t = 1 ?
(A) 50
(B) 100
(C) 40
(D) 25
5. [Online April 2019]
The position of a particle as a function of time t , is
given by x ( t ) = at + bt 2 − ct 3 where a , b and c are
constants. When the particle attains zero acceleration,
then its velocity will be
(A) a +
b2
4c
(B)
a+
b2
3c
(C) a +
b2
2c
(D) a +
b2
c
(B)
(C) 0.3 ms −1
(D) 0.1 ms −1
(B)
(C) y 2 = x 2 + constant
(D) xy = constant
9. [Online January 2019]
In a car race on straight road, car A takes a time t less
than car B at the finish and passes finishing point with
a speed v more than that of car B . Both the cars start
from rest and travel with constant acceleration a1 and
a2 respectively. Then v is equal to
(A)
a1 + a2
t
2
(B)
2 a1a2
t
a1 + a2
(C)
2 a1a2 t
(D)
a1a2 t
10. [Online January 2019]
The position co-ordinates of a particle moving in a
3-D coordinate system is given by x = a cos ( ωt ) ,
y = a sin ( ωt ) and z = aωt . The speed of the particle is
(A) 2aω
(C)
b 2τ
4
b 2τ
2
(D)
b 2τ
2
(C)
8. [Online January 2019]
A particle is moving with a velocity v = k yiˆ + xjˆ ,
where K is a constant. The general equation for its
path is
(
04_Kinematics 1_Part 4.indd 123
3aω
(D) aω
v(ms–1)
7. [Online April 2019]
A particle is moving with speed v = b x along positive x-axis . Calculate the speed of the particle at time
t = τ (assume that the particle is at origin at t = 0 ).
(B)
2aω
(B)
11. [Online January 2019]
A particle starts from the origin at time t = 0 and
moves along the positive x-axis . The graph of velocity with respect to time is shown in figure. What is the
position of the particle time t = 5 s ?
3
2
1
0
0.7 ms −1
(A) b 2τ
y = x 2 + constant
(A) y 2 = x + constant
6. [Online April 2019]
A bullet of mass 20 g has an initial speed of 1 ms −1 ,
just before it starts penetrating a mud wall of thickness 20 cm . If the wall offers a mean resistance of
2.5 × 10 −2 N , the speed of the bullet after emerging
from the other side of the wall is close to
(A) 0.4 ms −1
4.123
)
1 2 3 4 5 6 7 8 9 10
(A) 9 m
(B)
(C) 10 m
(D) 3 m
t(s)
6m
12. [Online January 2019]
A passenger train of length 60 m travels at a speed
of 80 km/hr . Another freight train of length 120 m
travels at a speed of 30 km/hr . The ratio of times
taken by the passenger train to completely cross the
freight train when : (i) they are moving in the same
direction and (ii) in the opposite directions is
(A)
25
11
(B)
5
2
(C)
11
5
(D)
3
2
11/28/2019 7:09:37 PM
4.124 JEE Advanced Physics: Mechanics – I
13. [2018]
All the graphs below are intended to represent the
same motion. One of them does it incorrectly. Pick
it up
P
d
(B) Distance
(A) Velocity
Time
Position
(C) Position
M
(A)
d
2
(B)
d
3
(C)
d
2
(D) d
17. [2017]
A body is thrown vertically upwards. Which one of the
following graphs correctly represent the velocity versus time?
(D) Velocity
Time
R
Q
Time
14. [Online 2018]
The velocity-time graphs of a car and a scooter are
shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and
(ii) the time at which the car will catch up with the
scooter are, respectively
(B) v
(A) v
t
t
(C) v
(D) v
G
15
0
O
5
(A) 112.5 m and 15 s
(C) 225.5 m and 10 s
C
D
10 15 20 25
Time (in s)
(B) 75 m
(D) 150 m
16. [Online 2018]
A man in a car at location Q on a straight highway
is moving with speed v . He decides to reach a point
P in a field at a distance d from the highway (point
M ) as shown in the figure. Speed of the car in the
field is half to that on the highway. What should be
the distance RM, so that the time taken to reach P is
minimum?
04_Kinematics 1_Part 4.indd 124
(B)
(A)
(B) 337.5 m and 25 s
(D) 112.5 m and 22.5 s
15. [Online 2018]
An automobile, travelling at 40 kmh −1 , can be stopped
at a distance of 40 m by applying brakes. If the same
automobile is travelling at 80 kmh −1 , the minimum
stopping distance, in metres, is (assume no skidding)
(A) 100 m
(C) 160 m
18. [Online 2017]
Which graph corresponds to an object moving with a
constant negative acceleration and a positive velocity?
Velocity
F
Distance
Distance
(C)
(D)
Velocity
E
Velocity
30
t
B
Velocity
Velocity (in ms–1)
t
A
45
Time
Time
19. [Online 2017]
The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of
the other rod has a roller that rolls along the floor in a
slot. As the roller goes back and forth, a 2 kg weight
moves up and down. If the roller is moving towards
right at a constant speed, the weight moves up with a
11/28/2019 7:09:42 PM
4.125
Chapter 4: Kinematics I
(C)
2 kg
240
Fixed pivot
F
8 12
x
Movable roller
3 th
of that of the roller when the
4
weight is 0.4 m above the ground.
(B) constant speed.
(C) decreasing speed.
(D) increasing speed.
(A)speed which is
20. [Online 2017]
A car is standing 200 m behind a bus, which is also at
rest. The two start moving at the same instant but with
different forward accelerations. The bus has acceleration 2 ms −2 and the car has acceleration 4 ms −2 . The
car will catch up with the bus after a time of
(A)
(B) 15 s
120 s
(C) 10 2 s
(D)
110 s
21. [2015]
Two stones are thrown up simultaneously from
the edge of a cliff 240 m high with initial speed of
10 ms −1 and 40 ms −1 respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the
first?
(Assume stones do not rebound after hitting the
ground and neglect air resistance, take g = 10 ms −2 )
(The figures are schematic and not drawn to scale)
(A)
240
(B)
( y2 – y1)m
240
t(s)
( y2 – y1)m
t(s)
12
t(s)
(A) gH = ( n − 2 ) u2
2
(C) gH = ( n − 2 ) u2
(B)
2 gH = n2u2
(D) 2 gH = nu2 ( n − 2 )
23. [2011]
An object moving with a speed of 6.25 ms −1 , is deceldv
erated at a rate given by
= −2.5 v , where v is the
dt
instantaneous speed. The time taken by the object, to
come to rest, would be
(A) 1 s
(B) 2 s
(C) 4 s
(D) 8 s
24. [2010]
A particle is moving with velocity v = K yiˆ + xjˆ ,
(
)
where K is a constant. The general equation for its
path is
(A) y 2 = x 2 + constant
(B)
y = x 2 + constant
(C) y 2 = x + constant
(D) xy = constant
25. [2009]
A particle has an initial velocity 3iˆ + 4 ˆj and an acceleration of 0.4iˆ + 0.3 ˆj . Its speed after 10 s is
iˆ
( y2 – y1)m
8 12
240
22. [2014]
From a tower of height H , a particle is thrown vertically upwards with a speed u . The time taken by the
particle, to hit the ground, is n times that taken by
it to reach the highest point of its path. The relation
between H , u and n is
(A) 10 units
(B)
7 2 units
(D) 8.5 units
(C) 7 units
8 12
(D)
( y2 – y1)m
t(s)
ARCHIVE: JEE ADVANCED
Single Correct Choice Type Problems
In this section each question has four choices (A), (B), (C)
and (D), out of which ONLY ONE is correct.
1. [IIT-JEE 2005]
The given graph shows the variation of velocity with
displacement. Which one of the graph given below
correctly represents the variation of acceleration with
displacement?
04_Kinematics 1_Part 4.indd 125
v
v0
x0
t
11/28/2019 7:09:50 PM
4.126 JEE Advanced Physics: Mechanics – I
(A) a
(B) a
(B) v
(A) v
d
x
h
d
x
(C) v
(C) a
(D) a
d
h
h
x
2. [IIT-JEE 2004]
A small block slides without friction down an inclined
plane starting from rest. Let Sn be the distance travS
elled from time t = n − 1 to t = n. Then n is
Sn+1
2n + 1
2n − 1
(C)
2n − 1
2n + 1
(D)
2n
2n + 1
3. [IIT-JEE 2004]
A particle starts from rest. Its acceleration ( a ) versus time ( t ) is as shown in the figure. The maximum
speed of the particle will be
a
10 ms–2
11
t(s)
(A) 110 ms −1
(B)
(C) 550 ms −1
(D) 660 ms −1
m
(B)
A
0
2n − 1
2n
5. [IIT-JEE 1999]
In 1.0 second a particle goes from point A to point B
moving in a semi-circle of radius 1.0 metre (figure).
The magnitude of the average velocity is
1.
(A)
B
(A) 3.14 ms −1
(B)
(C) 1.0 ms −1
(D) ZERO
2.0 ms −1
6. [IIT-JEE 1993]
A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0 . At this instant
of time, the horizontal component of its velocity is v .
A bead Q of same mass as P is ejected from A at
t = 0 along the horizontal string AB , with a speed
v . Friction between the bead and the string may be
neglected. Let tP and tQ be the respective times taken
by P and Q to reach the point B , then
55 ms −1
4. [IIT-JEE 2000]
A ball is dropped vertically from a height d above the
ground. It hits the ground and bounces up vertically
1
to a height d . Neglecting subsequent motion and
2
air resistance, its velocity v varies with the height h
above the ground as
04_Kinematics 1_Part 4.indd 126
(D) v
d
x
h
Q
A
B
P
C
(A) tP < tQ
(B)
tP = tQ
(C) tP > tQ
(D)
length of arc ACB
tP
=
tQ length of chord AB
11/28/2019 7:09:58 PM
Chapter 4: Kinematics I
7. [IIT-JEE 1988]
A boat which has a speed of 5 kmh −1 in still water
crosses a river of width 1 km along the shortest possible path in 15 minute. The velocity of the river water
in kmh −1 is
(A) 1
(B)
(C) 4
(D)
3
41
8. [IIT-JEE 1983]
A river is flowing from west to east at a speed of 5 m
per minute. A man on the south bank of the river,
capable of swimming at 10 m per minute in still water
wants to swim across the river to a point directly opposite in the shortest time. He should then swim in a
direction
(B) 30° east of north
(A) 60° west of north
(C) 30° west of north
(D) 60° east of north
9. [IIT-JEE 1982]
In the arrangement shown in the figure, the ends P
and Q of an unstretchable string move downwards
with uniform speed U . Pulleys A and B are fixed.
Mass M moves upwards with a speed
A
B
θ θ
P
(A) 2U cos θ
(C)
2U
cos θ
Q
M
(B)
U
cos θ
(D) U cos θ
10. [IIT-JEE 1982]
A particle is moving Eastwards with a velocity of
5 ms −1 . In 10 s , the velocity changes to 5 ms −1
Northwards. The average acceleration in this time is
(A) zero
(B)
1
ms −2 towards North-East
2
(C)
1
ms −2 towards North-West
2
(D)
1
ms −2 towards North
2
Multiple Correct Choice Type Problems
In this section each question has four choices (A), (B), (C)
and (D), out of which ONE OR MORE is/are correct.
04_Kinematics 1_Part 4.indd 127
4.127
1. [IIT-JEE 1999]
The co-ordinates of a particle moving in a plane are
given by x = a cos ( pt ) and y = b sin ( pt ) where a,
b ( < a ) and p are positive constants of appropriate
dimensions. Then
(A) the path of the particle is an ellipse.
(B)the velocity and acceleration of the particle are
π
.
normal to each other at t =
2p
(C)the acceleration of the particle is always directed
towards the focus.
(D)the distance travelled by the particle in time interπ
is a .
val t = 0 to t =
2p
2. [IIT-JEE 1993]
A particle of mass m moves on the x-axis as follows:
it starts from rest at t = 0 from the point x = 0 and
comes to rest at t = 1 at the point x = 1 . No other
information is available about its motion at intermediate time (0 < t < 1). If α denotes the instantaneous
acceleration of the particle, then
α cannot remain positive for all t in the interval
(A)
0≤t≤1
(B) α cannot exceed 2 at any point in its path
(C)α must be ≥ 4 at some point or points in its
path
α must change sign during the motion but no
(D)
other assertion can be made with the information
given
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical
value obtained after series of calculations based on the data
provided in the question(s).
1. [JEE (Advanced) 2014]
A rocket is moving in a gravity free space with a
constant acceleration of 2 ms −2 along +x direction
(shown in figure). The length of a chamber inside the
rocket is 4 m . A ball is thrown from the left end of
the chamber in +x direction with a speed of 0.3 ms −1
relative to the rocket. At the same time, another ball is
thrown in −x direction with a speed of 0.2 ms −1 from
its right end relative to the rocket. The time in seconds
when the two balls hit each.
a = 2 ms–2
x
0.3 ms–1
0.2 ms–1
4m
11/28/2019 7:10:05 PM
4.128 JEE Advanced Physics: Mechanics – I
2. [JEE (Advanced) 2014]
Airplanes A and B are flying with constant velocity
in the same vertical plane at angles 30° and 60° with
respect to the horizontal respectively as shown in figure. The speed of A is 100 3 ms −1 . At time t = 0 s ,
an observer in A finds B at a distance of 500 m .
This observer sees B moving with a constant velocity
­perpendicular to the line of motion of A. If at t = t0 ,
A just escapes being hit by B , t0 in seconds is
A
30°
B
60°
Assertion and Reasoning Type Problems
This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY
04_Kinematics 1_Part 4.indd 128
ONE is correct. Each question contains STATEMENT 1 and
STATEMENT 2. You have to mark your answer as
Bubble (A) I f both statements are TRUE and STATEMENT
2 is the correct explanation of STATEMENT 1.
Bubble (B) I f both statements are TRUE but STATEMENT 2
is not the correct explanation of STATEMENT 1.
Bubble (C) I f STATEMENT 1 is TRUE and STATEMENT 2
is FALSE.
Bubble (D) If STATEMENT 1 is FALSE but STATEMENT
2 is TRUE.
1. [JEE (Advanced) 2008]
Statement-I: For an observer looking out through
the window of a fast moving train, the nearby objects
appear to move in the opposite direction to the train,
while the distant objects appear to be stationary.
Statement-II: If the observer and the object are moving at velocities v1 and v2 , respectively with reference
to a laboratory frame, the velocity of the object with
respect to the observer is v2 − v1 .
11/28/2019 7:10:07 PM
Chapter 4: Kinematics I
4.129
Answer Keys—Test Your Concepts and Practice Exercises
Test Your Concepts-I (Based on
Displacement, Velocity, Acceleration,
Average Speed and Velocity)
1. 32 ms −1 , 67 m, 66 m
as
2
2.
3. 2 ms −1
4. 30 m, 15 ms −1
5. (a) ms −2 , ms −3
(b) 1 s
(c) 82 m, −80 m
(d) at t = 0 , v = 0 and a = 6 ms −2
at t = 1 s , v = 0 and a = −6 ms −2
at t = 2 s , v = −12 ms −1 and a = −18 ms −2
at t = 3 s , v = −36 ms −1 and a = −30 ms −2
at t = 4 s , v = −72 ms −1 and a = −42 ms −2
6. 135 ms −1 , 42 ms −2
7. 20.6 ms −1 , 76°
8. 0.222 ms −1 , 2.22 ms −1
9. (a) 14.125 m
(b) 1.75 ms −1 , 4.03 ms −1
10. −27 m , 69 m
11. −30 ms −2
Test Your Concepts-II
(Based on Constant Acceleration)
13. (a) 80 s
(b) 90 kmh–1
(c) 37 s
14. (a) 33.7 min
(b) 44.53 min
15. 24.5 s, 600 m
16. 89 m
17.
3
3
h,
h , No overtaking
4
2
18. vav =
3v0 ( v1 + v2 )
4v0 + v1 + v2
19. 2.27 s
20. 2280 m, Car A
21. 22.3 ms −1 , 1500 m
22. 20.6 s
23. (a) 15.8 s
(b) 391 m
(c) 29.5 ms −1
24. 4 ms–1 opposite to the direction of motion of the train.
25. 0.2 ms −2 , 0.8 ms −1
Test Your Concepts-III
(Based on Variable Acceleration)
5
1. 5 log e ⎛⎜ ⎞⎟ s
⎝ 4⎠
2. v = ± 2s
3. x =
v0 (
1 − e − kt ) , a = − kv0 e − kt
k
αβ ⎞
2. (a) ⎛⎜
t
⎝ α + β ⎟⎠
4. s = 2v0
1 αβ ⎞ 2
(b) ⎛⎜
t
2 ⎝ α + β ⎟⎠
6. v = v02 + 2kt
3. 3x
6. (a) 5 ms −1
(b) 1.67 ms −2
(c) 7.5 m
8. 0.74 s, 6.2 ms −2
9. 3.45t0
10. 8 m
11.
3
5
12. 4 × 10 −4 s , 2 × 107 ms −2
04_Kinematics 1_Part 4.indd 129
5. ±2 6 ms −1 , 4 m, ±4 2 m
7. v =
A(
1 − e − Bt )
B
8. v = −
9. (a)
16
( 8t + 1 )2
, a=
256
( 8t + 1 )3
5
ms −1
6
(b) −
55
ms −1
6
10. v = 5 ( e t − 1 ) , x = 5 ( e t − t − 1 )
11/28/2019 7:10:16 PM
4.130 JEE Advanced Physics: Mechanics – I
11. 80 kms −2 , 6.93 ms
12.
( 4iˆ + 8 ˆj ) ms−1
⎛ g + kv02 ⎞
1
13. hmax =
log e ⎜
g ⎟⎠
2k
⎝
14. −
14. 45 m
15. (a) 75 m
(b) 4.1 s
16. 14.7 ms −1 , 19.6 metre
17. 1.5 s
Test Your Concepts-VI
(Based on Planar Motion)
20
ms -2
3
4
15. v = ( 6t 2 − 2t 3 2 ) , x = ⎛⎜ 2t 3 − t 5 2 + 15 ⎞⎟
⎝
⎠
5
1. (a) 48 m
(b) 16 2 ms −1
Test Your Concepts-IV (Based on Graph)
2. 4 m, 37.8 ms −2
4. 114 m
5. 400 m
8. 46.47 ms −1
3. kωt
4. (a) y = x − α
10. 12.7 ms −1 , 22.8 ms −1 , 36.1 ms −1
11. (a) v = t 2 − 2t
(b) 6.67 m
12. 10 ms
7.
gt
2
4. 10 50 ms −1
7.
8.
gn2 ⎛ gn − 2u ⎞
8 ⎜⎝ gn − u ⎟⎠
2
u0
1+
cu02
mg
9. 1.26 × 10 ms
−2
10. (a) 216,000 metre
(b) 447.8 sec
11. 5 13 ms
−1
(a) 16.25 m
(b) 1.8 s
12. 73.302 metre
v0 t0
+
g 2
04_Kinematics 1_Part 4.indd 130
8. 32 ms −1
1. (a) 40 s
(b) 80 m
2. 60°
4. (a) 28°
(b) 1.48 hour
5. 3 kmhr −1
6. (a) 0.7 s
(b) 1.3 m
7. 2v sin ( 20° )
8. 5 2 km , 15 minutes
3
13.
( 10iˆ + 7 ˆj ) ms−1 , ( 12iˆ + 10 ˆj ) m
Test Your Concepts-VII
(Based on Relative Velocity)
2 g0 R
3. v =
2
(b) v = k 1 + ( 1 − 2α t ) , a = − ( 2α k ) ˆj
6. 201.25 ms −1 , 405 ms −2
−1
Test Your Concepts-V
(Based on Motion Under Gravity)
2.
x2
k
9. 2 5 ms −2 at an angle of α = tan −1 ( 2 ) NW
10. 3200 m
11. a =
12.
2u − gt
t
u
, at 45° perpendicular to stream flow
2
13. 8 ms −1 , 12°
3
14. (a) He should row at an angle 90° + sin −1 ⎛⎜ ⎞⎟ upstream.
⎝ 5⎠
­
(b) 160 s
11/28/2019 7:10:23 PM
Chapter 4: Kinematics I
15. (a) 3.65 s, 12.3 m
(b) 10 minute
(b) 19.8 ms −1
16. (a) 0.8 ms
(c)
−1
2
km
3
19. tan −1 ( 3 )
(b) 1.79 ms −1
(c) 120° upstream
18. (a)
4.131
20. Infinite,
10
minute
sin θ
3d
2
Single Correct Choice Type Questions
1. B
2. D
3. C
4. D
5. A
6. B
7. B
8. B
9. C
10. A
11. D
12. C
13. B
14. A
15. C
16. C
17. C
18. C
19. A
20. D
21. B
22. C
23. A
24. A
25. B
26. C
27. D
28. C
29. D
30. D
31. C
32. B
33. D
34. D
35. D
36. C
37. A
38. A
39. D
40. C
41. C
42. A
43. B
44. C
45. C
46. C
47. A
48. D
49. B
50. C
51. B
52. D
53. B
54. A
55. D
56. C
57. B
58. A
59. C
60. D
61. A
62. A
63. C
64. A
65. A
66. D
67. D
68. C
69. C
70. C
71. D
72. A
73. D
74. C
75. B
76. D
77. B
78. B
79. C
80. D
81. A
82. C
83. B
84. C
85. B
86. D
87. C
88. B
89. A
90. C
98. B
99. D
100. C
91. A
92. A
93. A
94. D
95. D
96. B
97. C
101. B
102. B
103. A
104. C
105. B
106. A
107. C
Multiple Correct Choice Type Questions
1. A, C, D
2. A, D
3. A, B, C
4. C, D
5. C, D
6. A, C, D
7. B, C
8. A, B
9. A, B, D
10. A, B, D
11. B, C
12. A, C
13. B, C
14. A, D
15. A, B, C, D
16. B, D
17. A, B, C, D
18. A, B, D
19. A, B, C
20. B, C, D
21. C, D
22. B, C, D
23. B, D
24. A, B, C, D
25. A, C, D
26. A, B, C
27. A, B, C
28. B, C
29. A, B, D
30. A, D
31. A, C
32. B, C, D
33. A, B, D
34. A, C
35. A, B, C
36. C, D
37. B, C, D
38. A, C
39. C, D
Reasoning Based Questions
1. D
2. D
11. D
12. C
3. A
4. C
5. D
6. A
7. D
8. A
9. A
10. D
Linked Comprehension Type Questions
1. B
2. D
3. C
4. B
5. C
6. B
7. B
8. D
9. B
10. D
11. A
12. C
13. D
14. C
15. A
16. B
17. B
18. A
19. D
20. D
21. B
22. C
23. B
24. A
25. A
26. B
27. C
28. A
29. C
30. A
04_Kinematics 1_Part 4.indd 131
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4.132 JEE Advanced Physics: Mechanics – I
31. C
32. C
33. C
34. D
35. D
36. B
37. A
38. D
39. A
40. B
41. C
42. B
43. C
44. D
45. A
46. A
47. C
48. B
49. C
50. B
51. C
52. C
53. A
Matrix Match/Column Match Type Questions
1. A → (s, u)
B → (s, t)
C → (s, u)
D → (s, t)
2. A → (q)
B → (r)
C → (s)
D → (p)
3. A → (q)
B → (r)
C → (p, s)
D → (t)
4. A → (r, t)
B → (p)
C → (q)
D → (s)
5. A → (t)
B → (s)
C → (p)
D → (q)
6. A → (q)
B → (r)
C → (p, q)
D → (s)
7. A → (r)
B → (p)
C → (q)
D → (p)
8. A → (s)
B → (r)
C → (p)
D → (q)
9. A → (r)
B → (p)
C → (r)
D → (s)
10. A → (r)
B → (p)
C → (s)
D → (q)
11. A → (p)
B → (q)
C → (s)
D → (r)
12. A → (q)
B → (p, q)
C → (p, r)
D → (r, s)
13. A → (r)
B → (p)
C → (r)
D → (r)
14. A → (r)
B → (q)
C → (s)
D → (p)
15. A → (r)
B → (s)
C → (p)
D → (q)
16. A → (r)
B → (q)
C → (s)
D → (p)
17. A → (r, s)
B → (r, s)
C → (p, q, r, t)
D → (p, q, r, t)
18. A → (q, s)
B → (r)
C → (p)
D → (s)
19. A → (r)
B → (s)
C → (p)
D → (q)
20. A → (s)
B → (r)
C → (q)
D → (p)
21. A → (p, r)
B → (p, r)
C → (p, q)
D → (p, q)
22. A → (p, q)
B → (p, q)
C → (r)
D → (p, q)
23. A → (q)
B → (r)
C → (s)
D → (p)
24. A → (p, r)
B → (q, r)
C → (q)
D → (q, s)
25. A → (q)
B → (r)
C → (r)
D → (p, r)
E → (q, r)
F → (s, u)
G → (p, s)
E → (r)
Integer/Numerical Answer Type Questions
1. 3
2. 15
3. 54
4. 1
5. (a) 2 (b) 2 (c) 1, 1 (d) 82
6. 517, 616
7. (a) 1000 (b) 100
8. 300
9. 2
10. 100, 20
11. 1
12. 10
13. 1
14. 2500
15. 2
16. 14, 250
17. 9
18. 36
19. 5, 200
20. 13, 13
21. (a) 60, 6 (b) 10.8, 9
22. 2
23. 5
24. 3
25. 60
26. 5
27. 4
28. 9
29. 1
30. 2
31. 1
32. 5
33. 6
34. 5.6
35. 10
36. 3
37. 4
38. 1
39. 1
40. 4
41. 1
42. 2
43. 1
44. 2
45. 4
46. 1
04_Kinematics 1_Part 4.indd 132
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Chapter 4: Kinematics I
4.133
ARCHIVE: JEE MAIN
1. D
2. C
3. D
4. A
5. B
6. B
7. C
8. C
9. D
10. B
11. A
12. C
13. B
14. D
15. C
16. B
17. C
18. A
19. C
20. C
21. A
22. D
23. B
24. A
25. B
6. A
7. B
8. C
9. B
10. C
ARCHIVE: JEE ADVANCED
Single Correct Choice Type Problems
1. A
2. C
3. B
4. A
5. B
Multiple Correct Choice Type Problems
1. A, B, C
2. A, D
Integer/Numerical Answer Type Questions
1. 2 or 8
2. 5
Assertion and Reasoning Type Problems
1. B
04_Kinematics 1_Part 4.indd 133
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This page is intentionally left blank
04_Kinematics 1_Part 4.indd 134
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CHAPTER
5
Kinematics II
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Curvilinear Motion and Angular Parameters (g) Oblique Projectile and Properties
(b) Angular, Centripetal and Tangential
(h) Relative Motion between two Projectiles
Acceleration
(i) Condition of Collision between two
(c) Kinematics of Circular Motion
Projectiles
(d) Motion of Particle in a Curved Track
( j) Motion of Projectile Up and Down an
(e) Radius of Curvature
Inclined Plane
(f) Horizontal Projectile and Properties
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
and Advanced) are also given.
CURVILINEAR MOTION
INTRODUCTION
The linear displacement ( x ), velocity ( v ) and accel
eration ( a ) are vectors. Correspondingly, the angular counterparts may also be vectors, because in
addition to the magnitude, we must also specify a
direction for them, namely, the direction of Axis of
Rotation in space. Let us first have our discussion for
θ , which commonly is called Angular Displacement
but is not. It should be called as angle traversed or
just finite angle, because it is not a vector. So, finite
angle θ is not a vector, because it does not add
like vectors, as Illustrated by the successive rotation technique for the book, where we observe that
( Rotation )1 + ( Rotation )2 ≠ ( Rotation )2 + ( Rotation )1 .
N
N
θ1
θ 2 90°
90°
S
θ1 + θ2
S
N
N
θ2
90°
S
θ1
90°
θ2 + θ1
S
05_Kinematics 2_Part 1.indd 1
11/28/2019 7:15:15 PM
5.2 JEE Advanced Physics: Mechanics – I
A book rotated q1 (90° as shown about an axis at right
angles to the page) and then q2 (90° as shown about a
north-south axis) has a different final orientation that
if rotated first through q2 and then q1. This property is
called the non-commutivity of finite angles under addition: q1 + q2 ≠ q2 + q1.
This non-commutative property associated with
finite rotation provides us a solid platform for calling
it a non-vector. In the first case, we rotate the book
π
through two successive rotations of
each and we
2
observe that θ + θ ≠ θ + θ .
1
2
2
ANGULAR DISPLACEMENT, ANGULAR
VELOCITY, ANGULAR AND CENTRIPETAL
ACCELERATION
From the above discussion, we have observed that,
the finite angle is not a vector, so cannot be called
as Angular Displacement. However, infinitesimal
angle dθ is a vector and hence is called as Angular
Displacement.
1
In the other case, if we rotate the book through
successive rotations of 20° , then too θ1 + θ 2 would
still differ from θ 2 + θ1 , but the difference would be
smaller. As a matter of fact, as the two rotations are
made smaller and smaller, the difference between the
two sums disappears rapidly. So, if the infinitesimal
angles are taken then we observe that the order of
addition no longer affects the results. Hence infinitesimal angular displacements are vector quantities.
N
θ1
θ2
20°
S
θ1 + θ2
S
N
N
θ1
θ2
20°
20°
S
S
θ1 + θ2
The lower group repeats the experiment for 20° displacements. We see here that q1 + q2 ≅ q2 + q1. Also we observe
that when q1, q2 → 0, the final positions approach each
other. Finite angles under addition tend to commute as
the angles become very small. Infinitesimal angles do
commute under addition, making it possible to treat
them as vectors.
05_Kinematics 2_Part 1.indd 2
r + dr ≅ r + dl
r = r + dr = r + dl = r
O
r dθ
r + dl
A dl
B
Consider a particle moving in a circle of radius r .
Let the particle go from point A to B such that
Arc AB = dl. If dθ is the angle subtended by the arc
at the centre of the circle, then dl = rdθ . Since dθ is
very small, so we can also say
= dl = dr
AB
N
20°
Axis of rotation
So, in triangle OAB, we can have
r + dr = r + dl
Since, both OA and OB are the radius of the same
circle, so
OA = OB = r r = r + dr = r + dl = r …(1)
The direction of dθ is given by Right Hand Thumb
Rule, according to which, “curl the fingers of Right
Hand in the sense of rotation, then thumb gives the
direction of dθ (angular displacement), ω (angular
velocity), α (angular acceleration), τ (torque) and L
(angular momentum)”.
So, in magnitude we have
dl = rdθ
11/28/2019 7:15:18 PM
Chapter 5: Kinematics II
ω av =
5.3
Δθ θ 2 − θ1
=
Δt
t2 − t1
In magnitude,
s
v ⊥ v sinα dθ
=
=
l
l
dt
ω in (rpm)
ω in (rph)
π
×2
60
π
× 2
3600
π
180
θ
⊥ =
P
l
θ
A
where q is the angle made by AB at any instant with
a convenient reference line (generally horizontal).
(b) Furthermore, the angular velocity of a particle can
be different about different points. For that, let us
consider a particle moving on the circumference
of the circle, as shown.
P
ω in rads–1
×
α
v
θ v
v
{Please note, it is dividing, not differentiating}
⇒ v = ω × r …(3)
dθ
where ω =
= Angular Velocity of the particle
dt
dl dr
=
= Velocity of the particle
v=
dt dt
ω=
co
Dividing both sides of (1) by dt , we get
dl dθ =
×r
dt dt
(a) Angular velocity is always to be defined with
respect to the point from where the position vector of the point is drawn. Consider a particle having a velocity v, as shown. Let us calculate w of the
point P w.r.t. another arbitrarily chosen point A
(say) or with respect to the observer at A is
v
π
Angle between dθ and r is , so
2
⎧
⎫
⎛π⎞
⎨ ∵ sin ⎜⎝ ⎟⎠ = 1⎬
d = d = r dθ 2
⎩
⎭
Conceptual Note(s)
=
Direction of dθ , ω , α , τ and L given by THUMB
But vectorially, we observe that
dl = dθ × r …(2)
|
CW SENSE
v|
CCW SENSE
n
O
si
O
v = rω …(4)
The ω , angular velocity of the particle is always to
be measured in SI units of rads −1 . Generally, ω can
also be expressed as rpm (revolution per minute), rph
(­revolution per hour), degrees −1 .
A
ϕ
R
ϕ
R
θ
C
Convenient reference line
(HORIZONTAL)
ω in degrees–1
05_Kinematics 2_Part 1.indd 3
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5.4 JEE Advanced Physics: Mechanics – I
Then the angular velocity of point P w.r.t. the
observer at the centre of the circle C is
dθ
ωPC =
dt
Similarly, the angular velocity of the point P w.r.t.
observer at the circumference of circle at A is
dϕ
ωPA =
dt
In triangle ACP , since θ = 2ϕ
⇒
ωPA =
d ⎛ θ ⎞ 1 ⎛ dθ ⎞ 1
⎜ ⎟= ⎜
⎟ = ωPC
dt ⎝ 2 ⎠ 2 ⎝ dt ⎠ 2
⇒ ωPC = 2ωPA
(c) If two particles A and B are moving with velocities
v1 and v2 in the directions shown in figure. Then
dl
= rate of change of distance between the two
dt
particles
v2
v1
θ2
θ1
A
l
B
dl
= relative velocity between them along AB
dt
dl
= v2 cosθ2 − v1 cosθ1
⇒
dt
and ω r = ωBA = angular velocity of B with respect
to A or angular velocity of line AB, then
ω r = ωBA =
v2 sinθ2 − v1 sinθ1
l
Magnitude of the relative angular velocity
between them can also be given by
Relative velocity perpendicular to AB
ωr =
Distance between th
hem
Illustration 1
The line joining P and Q is of constant length l and
the velocities of P and Q are in directions inclined at
05_Kinematics 2_Part 1.indd 4
angles a and b respectively with PQ. If u is the velocity of P, what is the angular speed of PQ?
u
β
α
P
l
Q
Solution
Let v be the velocity of Q . Then u cos α = v cos β
Since l = constant
⇒
dl
=0
dt
⇒
v=
u cos α
cos β Now angular speed of line PQ is given by
ω=
Relative speed ⊥ to PQ
l
⎛ u cos α ⎞
sin β − u sin α
v sin β − u sin α ⎜⎝ cos β ⎟⎠
⇒ ω=
=
l
l
⇒ ω=
u ( tan β cos α − sin α )
l
Illustration 2
Two points are moving with uniform velocities u
and v along the perpendicular axes, OX and OY .
The motion is directed toward O , the origin. When
t = 0 , they are at a distance a and b respectively
from O .
(a) Calculate the angular velocity of the line joining
them at time t .
(b) Show that it is greatest, when t =
au + bv
u2 + v 2
Solution
(a) At t = 0 , the particles are at
respectively.
A
and B
Let at time t , the particles are at P and Q
respectively.
11/28/2019 7:15:26 PM
Chapter 5: Kinematics II
ANGULAR, CENTRIPETAL, TANGENTIAL
AND TOTAL ACCELERATION
Y
v
B
Q
b
u
θ
π –θ
O
a
P
X
A
From figure, OP = a − ut …(1)
and OQ = b − vt …(2)
Let θ be the angle which the line PQ makes
with the direction OX at time t. From ΔQOP .
tan ( π − θ ) =
OQ b − vt
=
OP a − ut
b − vt
a − ut
⇒
− tan θ =
⇒
⎛ b − vt ⎞
tan θ = ⎜
⎝ ut − a ⎟⎠
⇒
⎛ b − vt ⎞
θ = tan −1 ⎜
…(3)
⎝ ut − a ⎟⎠
Differentiating equation (3), we get
dθ
=
dt
1
⎛ b − vt ⎞
1+ ⎜
⎝ ut − a ⎟⎠
dθ
⇒ ω = dt =
2
×
( ut − a ) ( −v ) − ( b − vt ) u
( ut − a )2
av − bu
( a − ut )2 + ( b − vt )2
(b) ω will be maximum when denominator is
­minimum, because numerator is constant.
2
2
Let, ( Den ) = S = ( a − ut ) + ( b − vt ) . For it to be
MINIMUM, we have
r
dS
= 2 ( a − ut ) ( −u ) + 2 ( b − vt ) ( −v ) = 0
dt
⇒ t =
au + bv
u2 + v 2 It can be seen that
d2S
dt 2
or ω is maximum, at
t=
05_Kinematics 2_Part 1.indd 5
au + bv
u2 + v 2
5.5
> 0 . Hence, S is ­minimum,
Since we have now understood the concept of ω so,
let us now discuss about angular acceleration ( α ) .
Angular acceleration is the rate of change of angular
velocity. Mathematically
dω
α=
dt
The direction of α is again given by Right Hand
Thumb Rule, as discussed earlier.
If a be the acceleration of the particle, then
dv
a=
dt
Since v = ω × r
dv d = (ω × r )
⇒ a=
dt dt
d(
dB dA Using
A × B) = A ×
+
×B
dt
dt dt
dv dr dω =ω× +
×r
⇒ a=
dt
dt dt
⇒ a = ω × v + α × r …(1)
The term ω × v is the centripetal or radial accelera tion of the particle and α × r is the tangential acceleration of the particle. So,
aC = ω × v …(2)
aT = α × r …(3)
In magnitude, aC = vω and aT = rα
⇒
aC = vω = rω 2 =
v2
…(4)
r
We could have prove this vectorially, by using
v = ω × r
Substituting in (2), we get
= ω × (ω × r )
aC
Using A × ( B × C ) = ( A ⋅ C ) B − ( A ⋅ B ) C , we get
a = ( ω ⋅ r )ω − ( ω ⋅ ω ) r
C
Now, since ω ⊥ r , so ω ⋅ r = 0
⇒ aC = 0 − ω 2 r ∵ ω ⋅ω = ω2
{
}
11/28/2019 7:15:31 PM
5.6 JEE Advanced Physics: Mechanics – I
⇒
⇒
aC = −ω 2 r …(5)
{∵ r = r }
aC = aC = rω 2 Please note that from (5), we conclude that the centripetal acceleration is directed radially inwards (see
the negative sign justifying this).
Also, if we had been asked to give the expressions for
centripetal force and tangential force, then we would
have just multiplied the above expressions with the
mass of the particle, say m, to get
Fc = mac = m ( ω × v ) = − ω 2 r and FT = maT = m ( α × r )
Illustration 3
A solid body rotates with deceleration about a stationary axis with an angular deceleration α = k ω ,
where ω is its angular velocity and k is a positive
constant. Find the mean angular velocity of the body
averaged over the whole time of rotation if at the initial moment of time its angular velocity was equal to
ω0 .
Solution
⇒
α =k ω
dω
−
=k ω
dt
ω
Problem Solving Technique(s)
(a) For motion with uniform angular acceleration, we
can use the following equations
∫ ω = −∫ kdt
ω0
0
ω
⇒
( 2 ω ) ω = −kt
1
θ = ω0 t + αt 2
2
⇒
kt
ω0 − ω =
2
ω 2 − ω 02 = 2αθ
kt ⎞
⎛
⇒ ω = ⎜ ω 0 − ⎟ …(1)
⎝
2⎠
ω = ω0 + αt
⎛ ω + ω0 ⎞
t
θ=⎜
⎝ 2 ⎟⎠
where, q is the angle traversed (in radian) in time t
(second)
−1 )
(
w0 is the initial angular velocity in rads ,
w is the final angular velocity ( in rads −1 ) at time t
and
−2
a is the angular acceleration ( in rads ) .
(b) If a is not a constant, analyse the problem and feel
free to use
dω
dθ
dω
ω=
=ω
or α =
dt
dt
dθ
(c) The SI unit of q is radian, w is rads −1 and a is rads −2 .
2π
= 2π f , where T is the period of revolution
(d) ω =
T
1
and f is the frequency and f = .
T
(e) Here too, the graphs interpret analogous to their
fellows in Linear Kinematics, like
Slope in the q-t graph gives angular velocity (w).
Slope in the w-t graph gives angular acceleration (a).
Area under a curve in w-t graph gives the angle
traversed (q).
05_Kinematics 2_Part 1.indd 6
⇒
t
dω
0
2
The body will stop when
kt
=0
2
ω0 −
⇒
t=
2 ω0
k
Now average angular velocity over this time interval
is
2 ω0
k
ω =
∫
2 ω0
k
ω dt
0
2 ω0
k
∫ dt
0
=
∫
0
2
kt ⎞
⎛
⎜⎝ ω 0 − ⎟⎠ dt
2
2 ω0
k
=
ω0
3
∫ dt
0
Illustration 4
The speed ( v ) of a particle moving in a circle of
radius R varies with distance s as v = ks , where k
is a positive constant. Calculate the total acceleration
of the particle.
11/28/2019 7:15:35 PM
5.7
Chapter 5: Kinematics II
Solution
v = ks
Since the particle is moving in a circle, so total acceleration, a is
a = aC2 + aT2
where aC =
⇒
aC =
⇒
aC =
⇒
a=
k 2 s2
d
⎛ ds ⎞
and aT = ( ks ) = k ⎜ ⎟ = kv
⎝ dt ⎠
R
dt
{
k 2 s2
ds
and aT = k ( ks ) = k 2 s ∵
= v = ks
R
dt
R2
dθ ⎞
dθ ⎞
⎛
⎛
⇒iˆ v = − ⎜ 4 R cos θ sin θ ⎟ iˆ + ⎜ 2R cos ( 2θ ) ⎟ ˆj
⎝
⎝
⎠
dt
dt ⎠
Since
v2
dv
and aT =
R
dt
k 4 s4
dr
Since v =
dt
4 2
2
+ k s = k s 1+
}
s2
R2 dθ
=ω
dt
⇒iˆ v = −2Rω ⎡⎣ − sin ( 2θ ) iˆ + cos ( 2θ ) ˆj ⎤⎦ …(2)
⇒ v = 2Rω dv
Further, we know that a =
dt
2
ˆ
ˆ
⇒i a = 4 Rω cos ( 2θ ) i + sin ( 2θ ) ˆj (
⇒
)
a = 4 Rω 2 UNIT VECTORS ALONG THE RADIUS ( r̂ )
AND THE TANGENT ( t̂ )
Illustration 5
A particle P moves along a circle of radius R so that
its radius vector r , relative to the point O at the circumference rotates with constant angular velocity ω .
Find the magnitude of the velocity of the particle and
the direction of its total acceleration.
Consider a particle P moving in a circle of radius r
centred at origin O . The angular position of the particle at some instant is say θ . Let us here define two
unit vectors, one is r̂ (called radial unit vector) which
is along OP and the other is t̂ (called the tangential
unit vector) which is perpendicular to OP. Now, since
Solution
ˆ
rˆ = t = 1
We can write these two vectors as
In triangle OPC , we get from Lami’s Theorem
r
R
=
sin ( π − 2θ ) sin θ
⇒
r
R
=
2 sin θ cos θ sin θ VELOCITY AND ACCELERATION OF
PARTICLE IN CIRCULAR MOTION
y
P
θ
r
O
θ
R
rˆ = cos θ iˆ + sin θ ˆj and tˆ = − sin θ iˆ + cos θ ˆj
R
C
x
The position vector of particle P at the instant shown
in figure can be written as
r = OP = rrˆ
⇒iˆ r = r cos θ iˆ + sin θ ˆj
(
)
Y
r
t
⇒
θ
r = 2R cos θ …(1)
Further we observe that
iˆ r = ( r cos θ ) iˆ + ( r sin θ ) ˆj
⇒ iˆ r = ( 2R cos 2 θ ) iˆ + ( 2R sin θ cos θ ) ˆj 05_Kinematics 2_Part 1.indd 7
r
θ
O
P
x
11/28/2019 7:15:42 PM
5.8 JEE Advanced Physics: Mechanics – I
The velocity of the particle can be obtained by differ
entiating r with respect to time t So, we get
dr d ⎡
iˆ v =
=
r cos θ iˆ + sin θ ˆj ⎤⎦
dt dt ⎣
(
)
dθ ⎞
dθ ⎞ ⎤
⎡⎛
⎛
iˆ v = r ⎢ ⎜ − sin θ ⎟ iˆ + ⎜ cos θ ⎟ ˆj ⎥
⎝
dt ⎠
dt ⎠ ⎦
⎣⎝
{
}
dθ
⇒iˆ v = rω − sin θ iˆ + cos θ ˆj ∵
= ω …(1)
dt
⇒ v = ( rω ) tˆ (
)
Thus, we observe that velocity of the particle is rω
along t̂ or in tangential direction. Acceleration of the
particle can be obtained by differentiating equation
(1) with respect to time t. So, we get
dv d ⎡
iˆ a =
=
rω − sin θ iˆ + cos θ ˆj ⎤⎦
dt dt ⎣
(
)
(
)
⎡ d
− sin θ iˆ + cos θ ˆj +
⇒iˆ a = r ⎢ ω
⎣ dt
(
)
⎡
⎛ dθ ⎞ ˆ
⎛ dθ ⎞ ˆ ⎤
⎛ dω ⎞
⇒iˆ a = rω ⎢ − cos θ ⎜
⎟ t̂
⎟⎠ i − sin θ ⎜⎝
⎟⎠ j ⎥ + r ⎜⎝
⎝
dt
dt
dt ⎠ ⎣
⎦
⎛ dω ⎞ ˆ
⇒iˆ a = − rω 2 ⎡⎣ cos θ iˆ + sin θ ˆj ⎤⎦ + r ⎜
t
⎝ dt ⎟⎠
dv
a = − rω 2 rˆ + tˆ …(2)
dt
Thus, acceleration of a particle moving in a circle has
two components one is along t̂ (along tangent) and
the other along − r̂ (or towards centre). In equation
(2), the first term is called the tangential acceleration.
( at ) and the second term is called radial or centripetal acceleration ( ar ) . Thus,
dv
= rate of change of speed
dt
v2
= vω {∵ v = rω }
r
Here, the two components are mutually perpendicular. Therefore, net acceleration of the particle will be
and ar = ac = rω 2 =
a = ar2 + at2 = ac2 + aT2
a=
05_Kinematics 2_Part 1.indd 8
2
2
2
⎛ v2 ⎞
⎛ dv ⎞
= ⎜ ⎟ +⎜
⎝ r ⎠
⎝ dt ⎠
⎝ dt ⎟⎠
( rω 2 )2 + ⎛⎜ dv ⎞⎟
at = 0 and a = ar = rω 2
dv
= positive, i.e.,
dt
at is along t̂ or tangential acceleration of par
ticle is parallel to velocity v because v = rω tˆ and
dv
at = tˆ.
dt
dv
(c) In decelerated circular motion,
= negative and
dt
hence, tangential acceleration is anti-parallel to
velocity v.
(b) In accelerated circular motion,
The speed of a particle moving in a circle of radius
r = 2 m varies with time t as v = t 2 , where t is in
second and v in ms −1 . Find the radial, tangential and
net acceleration at t = 2 s .
Solution
2
⇒
at =
(a) In uniform circular motion, speed (v) of the
dv
= 0. Thus
­particle is constant, i.e.,
dt
Illustration 6
⎤
⎛ dω ⎞
− sin θ iˆ + cos θ ˆj ⎥
iˆ ⎜
⎝ dt ⎟⎠
⎦
Conceptual Note(s)
Linear speed of particle at t = 2 s is v = ( 2 ) = 4 ms −1
2
v2 ( 4 )
=
= 8 ms −2
r
2
The tangential acceleration is
Radial acceleration ar =
dv
= 2t
dt
So, tangential acceleration at t = 2 s is
at =
a = ( 2 )( 2 ) = 4 ms −2
t
The net acceleration of particle at t = 2 s is
a=
⇒
( aC )2 + ( aT )2 = ( 8 )2 + ( 4 )2
a = 80 ms −2 KINEMATICS OF MOTION OF PARTICLE
IN A CURVED TRACK
Consider a curved track, 1234, having portions 12, 23
and 34 on which points X, Y and Z are taken respectively. The particle moving in a curved track has
­
following accelerations.
11/28/2019 7:15:50 PM
Chapter 5: Kinematics II
(a) Centripetal acceleration aC, acting radially inwards.
(b) Tangential acceleration aT , acting tangentially.
So, at the points X and Z, on the curved track the
­particle has two accelerations aC and aT .
Z
aT
ac
a
3
Conceptual Note(s)
a
X
aT
At the point Y , r → ∞ , so aC → 0 , hence from 2 to 3,
the particle just follows a straight track 23 under the
influence of a single force aT .
So, we can say that for a particle moving in a curved
track, net acceleration is given by
a=
( rω ) + ( rα )2 a = r ω +α
As observed, any curved track/path can be assumed
to be made of a large number of circular arcs of variable radii. The radius of curvature at a point is the
radius of the circular arc that suitably fits on the curve
at that point.
v2
where v is the tangential velocity and
r
can be denoted by vT .
⇒
v2
aC = T
r 05_Kinematics 2_Part 1.indd 9
a = aT2 + aC2 = a2x + a2y {For 2D motion }
where aT = rate of change of speed =
⇒
aT =
⇒
aT =
d
dt
( v +v )=
2
x
2
y
2
Radius of Curvature
Since aC =
2 2
4
(a) On any curved path (not necessarily a circular
one) the acceleration of the particle has two components aT and aN in two mutually perpendicular
directions. Component of a along v is aT and
perpendicular to v is aN or aC. So, we have
{∵ aC ⊥ aT }
a = aC2 + aT2 ⇒
(read as v perpendicular) and aC = aP i.e., centripetal
acceleration equals the acceleration of the particle
parallel to the radius of the curve and hence aC can
also be denoted by aP .
(ac = 0 )
ac
vT2 v⊥2
=
aC aP
where vT = v⊥ i.e., tangential velocity equals the
velocity of the particle perpendicular to the radius
of the curve and hence vT can also be denoted by v⊥
2
O
⇒
r=
4
aT
Y
1
⇒
5.9
v x ax + v y ay
v 2x + v 2y
=
vx
dv d v
=
dt
dt
dv y
dv x
+ vy
dt
dt
v 2x + v 2y
a⋅v
v
⇒ aT = component of a along v
(b) If a and aT are known, aC can easily be found using
the relation
a = aC2 + aT2
⇒
aC = a2 − aT2 = a2x + a2y − aT2
⇒
2
2
⎛ dv y ⎞
⎛ dv ⎞
⎛ dv ⎞
aC = ⎜ x ⎟ + ⎜
−⎜ ⎟
⎟
⎝ dt ⎠
⎝ dt ⎠
⎝ dt ⎠
2
11/28/2019 7:15:55 PM
5.10 JEE Advanced Physics: Mechanics – I
ILLUSTRATION 7
A balloon starts rising from the earth’s surface. The
ascension rate is constant and equal to v0 . Due to
the wind, the balloon gathers the horizontal velocity
component vx = ky , where k is a constant and y is
the height of ascent. Find the dependence of the following quantities on y .
(a) The horizontal drift of the balloon x ( y ) .
(b) The total tangential acceleration of the balloon.
(c) The normal acceleration of the balloon.
This is the trajectory of the particle which is a
parabola.
(b) For finding the tangential and normal accelerations, we require an expression for the velocity as
a function of height y . So, we have
vy = v0 and vx = ky
⇒
v = vx2 + vy2 = v02 + k 2 y 2
Therefore tangential acceleration,
aT =
SOLUTION
(a) Since the balloon is ascending at a constant rate.
So,
dy
dt
⇒ dy = v0 dt
⇒ y = v0 t
⇒
dv
=
dt
aT =
v0 =
{∵ y = v0 t }
⇒
x=
v02
⇒
2
⎛ dvy ⎞
⎛ dvx ⎞
⎜⎝
⎟⎠ + ⎜⎝
⎟⎠
dt
dt
a=
ax2 + ay2 =
a=
dy
dvx
=k
= kv0
dt
dt
⎧ dvy
⎫
= 0⎬
⎨∵
dt
⎩
⎭
(c) To find the normal acceleration, since we know
that
⎛ t2 ⎞
x = kv0 ⎜ ⎟
⎝ 2⎠
v0 ⎛ y ⎞
2 ⎜⎝ v0 ⎟⎠
k2 y2
2
Integrating, we get
x=k
v02 + k 2 y 2
0
Now, the total acceleration a is, given by
dx
vx =
= ky
dt
dx = kydt = kv0 t dt
⇒
k 2 yv0
dy
=
v 2 + k 2 y 2 dt
k2y
1+
Also, we have
⇒
k2y
2
a 2 = aN
+ aT2
2
⇒
aN = a 2 − aT2 =
1 ky 2
2 v0
kv0
⎛ ky ⎞
1+ ⎜ ⎟
⎝ v0 ⎠
2
Test Your Concepts-I
Based on Curvilinear Motion
1. The speed of a particle moving in a plane is equal
to the magnitude of its instantaneous velocity,
v = v = v 2x + v 2y . Show that the rate of change
dv v ⋅ a
of speed can be expressed as
=
and use
dt
v
05_Kinematics 2_Part 1.indd 10
(Solutions on page H.143)
dv
this result to explain why
is equal to aT , the
dt
component of a that is parallel to v.
2. If a point moves along a circle with constant speed,
prove that its angular speed about any point on the
circle is half of that about the centre.
11/28/2019 7:16:02 PM
Chapter 5: Kinematics II
3. Calculate the angular speed of a particle which
moves in a circle of radius 0.25 m with a linear
speed of 2 ms −1.
4. A particle moves in a circle of radius 0.25 m at a
speed that uniformly increases. Find the angular
acceleration of particle if its speed changes from
2 ms −1 to 4 ms −1 in 4 s.
5. A car is travelling along a circular curve that has a
radius of 50 m. If its speed is 16 ms −1 and is increasing uniformly at 8 ms −2 , determine the magnitude
of its acceleration at this instant.
6. A boy whirls a stone in a horizontal circle of radius
0.5 m and at height 20 m above the level ground.
The string breaks, and the stone flies off horizontally to strike the ground after travelling a horizontal distance of 10 m. Calculate the magnitude of the
centripetal acceleration of the stone while in circular motion? Take g = 10 ms −2
7. Two particles A and B start from the origin O
and travel in opposite directions along the circular path at constant speeds v A = 0.7 ms −1
and vB = 1.5 ms −1, respectively. Determine the
time when they collide and the magnitude of
the acceleration of B just before this happens.
22 ⎞
⎛
⎜⎝ Take π = ⎟⎠
7
(
)
y
7m
vB
vA
B
A
O
x
8. Starting from rest, the motorcycle travels around
the circular path of radius 50 m, at a speed
t2
v=
ms −1, where t is in seconds. Determine the
5
05_Kinematics 2_Part 1.indd 11
5.11
magnitudes of the motorcycle’s velocity and acceleration at the instant t = 3 s.
ρ = 50 m
v
9. Determine the maximum constant speed a race
car can have if the acceleration of the car cannot
exceed 7.5 ms −2 while rounding a track having a
radius of curvature of 200 m.
10. An automobile is travelling on a horizontal circular
curve having a radius of 800 m. If the acceleration
of the automobile is 5 ms −2 , determine the constant speed at which the automobile is travelling.
11. A car travels along a horizontal circular curved road
that has a radius of 600 m. If the speed is uniformly
increased at a rate of 2000 kmh−2 , determine the
magnitude of the acceleration at the instant the
speed of the car is 60 kmh−1.
12. When designing a highway curve it is required that
cars travelling at a constant speed of 25 ms −1 must
not have an acceleration that exceeds 3 ms −2 .
Determine the minimum radius of curvature of the
curve.
13. At a given instant, a car travels along a circular curved road with a speed of 20 ms −1 while
decreasing its speed at the rate of 3 ms −2 . If the
magnitude of the car’s acceleration is 5 ms −2 ,
determine the radius of curvature of the road.
14. The wheel of a truck moving with velocity 5 ms −1
throws up mud particles from its rim. The diameter
of the wheel is 3 m. Find the maximum height from
ground to which a particle can reach? (Assume no
sliding between the wheel and road).
11/28/2019 7:16:04 PM
projectile motion
projectile motion: AN INTRODUCTION
Oblique Projectile
A particle launched with an initial velocity in any
arbitrary direction and then allowed to move freely
under the gravitational influence of the earth ( g ) is
called a projectile.
For studying the projectile motion the following
assumptions must be kept in mind.
In this, the projectile is given an initial velocity making an angle θ with the horizontal (or the vertical)
and it moves under the influence of gravity to follow
a parabolic path.
Assumption 1: The surface of the earth is more or
less flat.
Assumption 2: Acceleration due to gravity ( g ) has
a constant value throughout the entire motion.
Assumption 3: No air drag is present.
EXAMPLES:
(a) A cricket ball thrown by a bowler.
(b) A football kicked by the player.
(c) Food packet dropped from an airplane and many
more examples can be thought of as projectiles from
everyday life.
Conceptual Note(s)
The path followed by a projectile is called its
­trajectory, which is a parabola.
Types of Projectile Motion
u
g
u
g
u
(c)
(b)
g
(d)
Projectile on an Inclined Plane
In this, the projectile is given an initial velocity making an angle with the horizontal on an inclined plane
and it again moves under the influence of gravity to
follow a parabolic path.
u
g
g
u
(e)
(f)
We now discuss in detail the equation of trajectory,
velocity any point P and at particular instant and
position.
We shall study three kinds of projectile motion.
Horizontal Projectile
Horizontal Projectile
In this, the projectile is given an initial velocity
directed along the horizontal and then it moves under
the influence of gravity to follow a parabolic path.
u
Equation of Trajectory
Consider a point P(x , y ) at time t .
g
g
O
ux = u
x
y
uy = 0
x
(a)
y
05_Kinematics 2_Part 1.indd 12
P(x, y)
vx = ux = u
β
vy = gt
v
11/28/2019 7:16:06 PM
Chapter 5: Kinematics II
Horizontal Motion (HM)
Since acceleration due to gravity acts along the vertical and hence, has got no component along the horizontal i.e., ax = 0.
So, horizontal motion is a non-accelerated motion
with uniform velocity.
⇒
x = ux t = ut …(1)
Vertical Motion (VM)
Also ay = g and uy = 0
⇒
y = 0+
If b is the angle made by v with the horizontal, then
1 2
gt
2
1
⇒ y = gt 2 …(2)
2
x
From (1), we get t =
u
Substituting in (2), we get
⎛ g ⎞
y = ⎜ 2 ⎟ x2
⎝ 2u ⎠
which is the equation of a parabola.
tan β =
vy
vx
=
gt
u
If h is the distance of the ground from the point
of launch, T is the time taken to strike the ground
and R is the range of the projectile when it hits the
ground, then
h=
1 2
ay t
2
Since, y = uy t +
1 2
gT
2
⇒
T=
2h
and R = uT g
⇒
R=u
2h
g
Deviation Suffered by a Horizontal
Projectile in Time t
If gravity were absent, then the body launched with
initial velocity u would continue to move horizontally forever. However in the presence of gravity it
1 2
suffers a deviation, δ = y = gt . So,
2
Deviation = δ = y =
Velocity at Any Instant (t)
1 2 ⎛ g ⎞ 2
gt = ⎜ 2 ⎟ x
⎝ 2u ⎠
2
x = (ut)i
Horizontal Motion (HM)
x
Since horizontal motion is non-accelerated,
⇒
vx = ux = u
g
y = 1 gt2 j
2
Vertical Motion (VM)
Trajectory
Also, vy = uy + ay t
⇒
vy = 0 + gt
⇒
vy = gt
y
Illustration 8
A projectile is fired horizontally with a velocity of
98 ms −1 from the top of a hill 490 m high. Find
Since, we have
v = vx iˆ + vy ˆj
⇒
⇒
v = uiˆ + ( gt) ˆj
2
v = v = u + g 2t 2
05_Kinematics 2_Part 1.indd 13
5.13
(a) the time taken by the projectile to reach the
ground
(b) the distance of the target from the hill and
(c) the velocity with which the projectile hits the
ground. g = 9.8 ms −2 .
(
)
11/28/2019 7:16:12 PM
5.14 JEE Advanced Physics: Mechanics – I
Solution
Illustration 9
Here, it will be more convenient to choose x and y
directions as shown in figure. So, let the downward
direction be positive, then
Two particles move in a uniform gravitational field
with an acceleration g . At the initial moment the
particles were located at one point and moved with
velocities u1 = 3 ms −1 and u2 = 4 ms −1 horizontally
in opposite directions. Find the distance between the
particles when their velocity vectors become mutually perpendicular.
ux = 98 ms −1 , ax = 0 , uy = 0 and ay = g
O u = 98 ms
–1
x
Solution
y
P
A
β
x
vy
vx
v
(a) At A , y = 490 m . So, applying
⇒
⇒
y = uy t +
1 2
ay t
2
Let the velocities of the particles be at right angles to
each other at time t . Then
v1 = u1 + gt
⇒ v1 = 3iˆ + gtjˆ v 2 = u2 + gt
⇒ v 2 = −4iˆ + gtjˆ 1
490 = 0 + ( 9.8 ) t 2
2
t = 10 s
⇒
1 2
ax t
2
PA = ( 98 )( 10 ) + 0
⇒
PA = 980 m
O
(b) PA = x = ux t +
(c) vx = ux = 98 ms −1
⇒
vy = uy + ay t = 0 + ( 9.8 )( 10 ) = 98 ms −1
2
2
v = vx2 + vy2 = ( 98 ) + ( 98 ) = 98 2 ms −1
and tan β =
vy
vx
=
98
=1
98
v1
05_Kinematics 2_Part 1.indd 14
Path followed
Path followed
by 1
by 2
at time t
v2
For velocities to be mutually perpendicular, we have
v1 ⋅ v 2 = 0
⇒
−12 + g 2 t 2 = 0 ⇒
t=
12
g
2
=
2 3
s
g
Separation between the particles is
⇒ β = 45°
Thus, the projectile hits the ground with a
­velocity 98 2 ms −1 at an angle of β = 45° with
horizontal as shown in figure.
B
A
x = (u1 + u2 )t = 7
⇒
x=
14 3
m
g
2 3
g
11/28/2019 7:16:18 PM
Chapter 5: Kinematics II
5.15
Test Your Concepts-II
Based on Horizontal Projectile
1. A body is thrown horizontally from the top of a
tower and strikes the ground after three seconds at
an angle of 45° with the horizontal. Find the height
of the tower and the speed with which the body
was projected. Take g = 9.8 ms −2 .
2. With what minimum horizontal velocity u can
a boy throw a ball at A and have it just clear the
obstruction at B?
u
(Solutions on page H.144)
Calculate the horizontal launch velocity of the
(
)
rock. Take g = 10 ms −2 .
6. Calculate the minimum velocity u along the horizontal such that the ball just clears the point C.
Assume that the ball is launched by a man who
holds the ball at a distance 1 m above A. Also
find x, where the ball strikes the ground. Take
g = 9.8 ms −2 .
40 m
u
1m
A
A
3.9 m
C
B
B
36 m
6m
16.4 m
14.7 m
3. The velocity of the water jet discharging from the
orifice (small hole in the tank) can be obtained
from u = 2 gh , where h = 5 m is the depth of the
orifice from the free water surface. Determine the
time for a particle of water leaving the orifice to
reach point B and the horizontal distance x, where
it hits the surface. Take g = 10 ms −2 .
5m
A
u
B
2.5 m
x
4. Determine the horizontal velocity u of a tennis
ball launched from A at height of 2.25 m from the
ground so that it just clears the net of height 1 m
at B, a horizontal distance of 6.4 m from A. Also,
find the horizontal distance from the net, where
the ball strikes the ground. Take g = 10 ms −2 .
5. A rock is thrown horizontally from top of a tower
and hits the ground 4 s later. The line of sight from
top of tower to the point where the rock hits the
ground makes an angle of 30° with the horizontal.
05_Kinematics 2_Part 1.indd 15
D
x
7. Ball bearings of diameter 20 mm leave the horizontal with a velocity of magnitude u and fall through
the 60 mm diameter hole at a depth of 800 mm as
shown. Calculate the permissible range of u which
will enable the ball bearings to enter the hole. Take
the dotted positions to represent the limiting conditions. Take g = 10 ms −2
(
20 mm
)
120 mm
u
800 mm
60 mm
11/28/2019 7:16:20 PM
5.16 JEE Advanced Physics: Mechanics – I
OBLIQUE PROJECTILE
Velocity at Any Instant (t)
Let a projectile be launched with an initial velocity u
making an angle θ with horizontal then
Horizontal Motion (HM)
Since horizontal motion is non-accelerated. So,
vx = ux = u cos θ …(4)
ux = u cos θ and uy = u sin θ
Vertical Motion (VM)
y
Since,
g
u
uy
vy
v
β
P
vx = ux
H
y
v y = uy + ay t
θ
O
⇒
vy = 0
ux = u cos θ
⇒
x
ux
Equation of Trajectory
Since horizontal motion is non-accelerated
i.e., ax = 0. 1 2
ax t
2
Vertical Motion (VM)
vx
=
u sin θ − gt
…(7)
u cos θ
Vertical Motion (VM)
vy2 − uy2 = 2 ay y
1
y = ( u sin θ ) t + ( − g )t 2 {∵ ay = − g }
2
1
⇒ y = ( u sin θ ) t − gt 2 …(2)
2
x
From (1), t =
and put in (2), we get
u cos θ
⎛ x ⎞ 1 ⎛ x ⎞
y = (u sin θ ) ⎜
− g
⎝ u cos θ ⎟⎠ 2 ⎜⎝ u cos θ ⎟⎠
gx 2
2u2 cos 2 θ
2
…(3)
which is the equation of a parabola
Since horizontal motion is a non-accelerated motion,
so
vx = ux = u cos θ …(8)
1
y = uy t + ay t 2
2
If we take upwards as positive, then
05_Kinematics 2_Part 1.indd 16
vy
Horizontal Motion (HM)
x = ( u cos θ ) t …(1)
y = x tan θ −
tan β =
Velocity at Any Position at Vertical
Height (h)
Also,
⇒
v = u2 + g 2 t 2 − 2ugt sin θ
If β is the angle made by v with x-axis , then
Horizontal Motion (HM)
⇒
vy = u sin θ − gt …(5)
From figure, we have
v = vx iˆ + vy ˆj
⇒ iˆ v = ( u cos θ ) iˆ + ( u sin θ − gt ) ˆj …(6)
⇒
Consider a point P(x , y ) at time t.
x = ux t +
vy = u sin θ + ( − g )t
⇒
v 2 sin 2 β = u2 sin 2 θ + 2( − g )h ⇒
vy = u2 sin 2 θ − 2 gh …(9)
Since,
⇒ v = vx iˆ + vy ˆj
ˆ
ˆ
⇒ i v = ( u cos θ )i + ( u2 sin 2 θ − 2 gh ) ˆj …(10)
⇒
v = v = u2 − 2 gh
If β is the angle made by v with x-axis , then
tan β =
vy
vx
=
u2 sin 2 θ − 2 gh
u cos θ
…(11)
11/28/2019 7:16:27 PM
Chapter 5: Kinematics II
Problem Solving Technique(s)
⇒
Projectile motion is a two dimensional motion with
constant acceleration (g). So, we can use
1
v = u + at, s = ut + at 2 , etc. in projectile motion as
2
well. Here
iˆ u = u cos α iˆ + u sinα ˆj and a = − gjˆ
Now, suppose we want to find velocity at time t .
iˆ v = u + at = ( u cos α iˆ + u sinα ˆj ) − gtjˆ
⇒ iˆ v = u cos α iˆ + ( u sinα − gt ) ˆj
Similarly displacement at time t will be,
1 2
s = ut + at
2
2u sin θ
g
tascent = tdescent
⇒
tascent = tdescent =
Range (R)
Maximum horizontal distance travelled is called
Range.
At t = T , x = R
⇒
u
O
T u sin θ
=
2
g
1 2
ax t
2
and ax = 0 {∵ horizontal motion is non-accelerated}
g
α
{∵T ≠ 0 }
Further, T = tascent + tdescent and since no air drag
exists, so,
Since x = ux t +
y
x
1
iˆ s = ( u cos α iˆ + u sinα ˆj ) t − gt 2 ˆj
2
T=
5.17
1
⎛
⎞
iˆ s = ut cos α iˆ + ⎜ ut sinα − gt 2 ⎟ ˆj
⎝
⎠
2
⇒
R = (u cos θ )T
2
R = (u cos θ )(u sin θ )
g
⇒
R=
Vertical
⎞
2 ⎛ Horizontal ⎞ ⎛
⎜ Component of ⎟ ⎜ Component of ⎟
g ⎜⎝ Initial Velocity ⎟⎠ ⎜⎝ Initial Velocity ⎟⎠
⇒
R=
u2 sin( 2θ )
g
Maximum Height (H )
Maximum vertical displacement of projection is called
Maximum Height. At maximum height vy = 0.
Since, vy2 − uy2 = 2 ay y
⇒
0 2 − (u sin θ )2 = 2( − g )H ⇒
H=
u2 sin 2 θ
2g
Time of Flight (T )
Time taken by projectile to strike the ground. At t = T ,
y = 0 (because projectile returns to the ground). Since
1
y = uy t + ay t 2
2
⇒
1
0 = (u sin θ )T + ( − g )T 2
2
05_Kinematics 2_Part 1.indd 17
Problem Solving Technique(s)
When the ball is projected from a body as per the
cases discussed.
Case-1: Body moving in same direction
Consider that a man is sitting in a trolley and the trolley is moving with velocity v in positive x direction as
shown. Let the man projects a ball in the direction of
trolley.
y
u
θ
v
x
11/28/2019 7:16:33 PM
5.18 JEE Advanced Physics: Mechanics – I
Case-4: Body moving down
Consider that a man is standing in a lift compartment
which is moving down with velocity v as shown. Let
the man project a ball in the direction shown.
In this case, we have
Horizontal component ux = u cosθ + v
Vertical component uy = u sinθ
⇒
R=
2
( u cosθ + v )( u sinθ )
g
y
Case-2: Body moving in opposite direction
Consider that a man is sitting in a trolley and the trolley is moving with velocity v in negative x direction as
shown. Let the man projects a ball in the direction of
trolley.
y
u
θ
v
x
In this case, we have
Horizontal component ux = u cosθ
u
Vertical component uy = u sinθ − v
θ
v
⇒
x
In this case, we have
Horizontal component ux = u cosθ − v
R=
2
( u cosθ ) ( u sinθ − v )
g
Deviation Suffered by an Oblique Projectile
in Time t
Vertical component uy = u sinθ
⇒
R=
2
( u cosθ − v )( u sinθ )
g
Case-3: Body moving up
Consider that a man is standing in a lift compartment
which is moving up with velocity v as shown. Let the
man project a ball in the direction shown.
Again we proceed with the same kind of thought process, where we assume gravity to be absent first. Then
the body launched with initial velocity would continue to move forever along the line OA. However
gravity is present, due to which it suffers a deviation
δ from its actual track (in the absence of gravity). In
triangle OAN.
y
tan θ =
u
v
AN y + δ
=
ON
x
⇒ δ = x tan θ − y …(1)
θ
Now since we know that y = x tan θ −
x
g
A
Vertical component uy = u sinθ + v
R=
δ
u
2
( u cosθ ) ( u sinθ + v )
g
AP = Deviation (δ )
PN = y
ON = x
P
y
θ
O
05_Kinematics 2_Part 1.indd 18
2u2 cos 2 θ
y
In this case, we have
Horizontal component ux = u cosθ
⇒
gx 2
x
N
x
11/28/2019 7:16:36 PM
Chapter 5: Kinematics II
⎛
⎞
gx 2
⇒ δ = ( x tan θ ) − ⎜ x tan θ − 2
2 ⎟
⎝
2u cos θ ⎠
The situation is shown in figure. In this case, we have
2
1 ⎛ x ⎞
g⎜
⎟
2 ⎝ u cos θ ⎠ Since, x = ( u cos θ ) t
⇒ δ=
⇒ δ=
5.19
y = +h , ux = u cosθ , uy = +u sinθ and a y = + g
For horizontal motion,
x = ( u cosθ ) t
For vertical motion,
1 2
gt
2
1
+h = ( u sinθ ) t + gt 2
2
Hence deviation suffered by an oblique projectile in
time t (or in terms of x , u and launch angle θ ) is
⇒
gt 2 + ( 2u sinθ ) t − 2h = 0
gx 2
1
δ = gt 2 = 2
2
2u cos 2 θ
⇒
t=
−u sinθ
u2 sin2 θ 2h
±
+
g
g
g2
g
O
Problem Solving Technique(s)
Case-1: Particle projected at an angle q above the
horizontal
u
O
θ
x
u
y
+
+
h
g
θ
+ve
+ve
h
Here negative root should be excluded otherwise t
would be negative.
Illustration 10
The situation is shown in figure. In this case, we have
y = −h , ux = u cosθ , uy = u sinθ and a y = − g
For horizontal motion,
x = ( u cosθ ) t
For vertical motion,
1 2
−h = ( u sinθ ) t − 2 gt
⇒
gt 2 − ( 2u sinθ ) t − 2h = 0
⇒
t=
u sinθ
u2 sin2 θ 2h
±
+
g
g
g2
Case-2: Particle projected at an angle q below the
horizontal
05_Kinematics 2_Part 1.indd 19
A stone is thrown from the top of a tower of height
50 m with a velocity of 30 ms −1 at an angle of 30°
above the horizontal. Find
(a) the time during which the stone will be in air,
(b) the distance from the tower base to where the
stone will hit the ground,
(c) the speed with which the stone will hit the
ground,
(d) the angle formed by the trajectory of the stone
with the horizontal at the point of hit.
Solution
The situation is shown in figure
(a) Horizontal component of velocity
ux = 30 cos ( 30° ) = 25.98 ms −1
11/28/2019 7:16:42 PM
5.20 JEE Advanced Physics: Mechanics – I
uy
the horizon. Another particle B is projected with the
same velocity u but in a downward direction exactly
­opposite to A. The two particles will strike the ground
at a distance x apart. Find x.
u = 30 ms–1
30°
ux
u sin α
50 m
u
α
u cos α
α
O
x
vy
u
vx
θ
v
Vertical component of velocity
vy = 30 sin ( 30° ) = 15 ms −1
Let t be the time taken by the stone to reach the
ground, i.e., the time during which the stone will
be in air. Taking the upward direction as p
­ ositive,
we have
{
1
× 10 × t 2 2
Solving for t , we get
−50 = 15t −
∵ h = ut +
1 2
gt
2
}
(b) The distance x, where the stone will hit the
ground
x = ux t = 25.98 × 5 = 104.90 m
(c) From figure, vx = ux = 25.98 ms −1
and vy = −15 + ( 10 × 5 ) = 35 ms −1
vx2 + vy2
⇒
v = ⎡⎣ ( 25.98 ) + ( 35 ) ⎦ = 43.6 ms
2
2⎤
−1
⎛ vy ⎞
(d) tan θ = ⎜
⎝ v ⎟⎠
x
⇒
35 ⎞
θ = tan ⎜
≅ 53°
⎝ 25.98 ⎟⎠
Solution
Taking upward direction as positive. Then
For particle A
Vertical displacement = −h (downwards)
−1 ⎛
and Acceleration = −g (downwards)
From a point at a height h above the horizontal
ground a particle A is projected with a velocity u
in an upward direction making an angle α with
1 2
gt1
2
⇒
− h = ( u sin α ) t1 −
⇒
1 2
gt1 − ( u sin α ) t1 − h = 0
2
⇒
gt12 − 2u sin α t1 − 2 h = 0 ⇒
t1 =
⇒
t=
2u sin α ± 4u2 sin 2 α + 8 gh
2g
u sin α ± u2 sin 2 α + 2 gh
g
If we use negative sign, then time t1 will be negative.
So, dropping the negative sign, we get
Illustration 11
05_Kinematics 2_Part 1.indd 20
x2
Vertical component of velocity = u sin α (upwards)
t=5s
Since v =
x1
t1 =
u sin α + u2 sin 2 α + 2 gh
g
11/28/2019 7:16:47 PM
Chapter 5: Kinematics II
For particle B
We have Vertical Displacement = −h (downwards)
Vertical Component of Velocity = −u sin α (downwards) and Acceleration = −g (downwards)
y
g
25 ms–1
1 2
gt2
2
⇒
− h = − ( u sin α ) t2 −
⇒
h = ( u sin α ) t2 +
⇒
gt22 + 2 ( u sin α ) t2 − 2 h = 0 5m
1 2
gt2
2
O
θ
x
2s
A
R
x
R–x
Solving, we get
2
⇒
t2 =
Greatest height reached is H =
2
−u sin α + u sin α + 2 gh
g
t2 is the time taken by the particle to reach the
ground
Now, distance travelled x = x1 − ( − x2 ) = x1 + x2
⇒
x = ( u cos α ) t1 + ( u cos α ) t2 ⇒
x = u cos α ( t1 + t2 )
⇒
⎛2 2
⎞
x = u cos α ⎜
u sin 2 α + 2 gh ⎟
⎝g
⎠
⇒
x=
⇒
H=
=
u2 sin 2 θ
2g
( 25 )2 ⎛⎜ 1 ⎞⎟
⎝ 4⎠
= 7.81 m
2 ( 10 )
Range = R =
2u cos α
u2 sin 2 α + 2 gh
g
5.21
2
u2 sin ( 2θ ) ( 25 ) sin ( 60 )
=
10
g
( 625 ) 3
20
= 31.25 3 m
⎛ 3⎞
( 2 ) = 25 3 m
Since x = ( u cos θ ) t = ( 25 ) ⎜
⎝ 2 ⎟⎠
So, ( R − x ) = 6.25 3 m = 10.8 m
Illustration 12
Illustration 13
A stone is projected from the ground with a velocity of 25 ms −1 . Two second later it just clears a wall
5 m high. Find the angle of projection of the stone,
the greatest height reached. How far beyond the wall
the stone will again hit the ground. g = 10 ms −2 .
Two seconds after the projection a projectile is travelling in a direction inclined at 30° to the horizon.
After one more second it is travelling horizontally.
Determine the magnitude and direction of its velocity.
Solution
At t = 2 s
1
Since y = uy t + ay t 2
2
1
( − g ) t2
2
⇒
5 = ( u sin θ ) t +
⇒
1
2
5 = ( 25 sin θ ) 2 − ( 10 ) ( 2 )
2
⇒
sin θ =
1
2
⇒ θ = 30° 05_Kinematics 2_Part 1.indd 21
Solution
v cos 30 = u cos θ …(1)
Also,
v y = uy + ay t
⇒
v sin 30 = u sin θ − 20 …(2)
After one more second, it is travelling horizontally
i.e., it is at the maximum height. So, time of ascent
is 3 s . Hence
11/28/2019 7:16:54 PM
5.22 JEE Advanced Physics: Mechanics – I
y
u sin θ
=3
g
⇒
u sin θ = 30 …(3)
A
u
20 ms–1
From (2) and (3), we get
45°
v
= 30 − 20
2
⇒
O
u cos θ = 10 3 …(4)
Squaring (3) and (4) and adding, we get
u2 = 900 + 300
⇒
u = 1200 ⇒
u = 20 3 ms −1 From (1), we get
10 3 = 20 3 cos θ
⇒ θ = 60° So, the particle was launched with an initial ­velocity
of 20 3 ms −1 making an angle of 60° with the
horizontal.
Illustration 14
A gun kept on a horizontal straight road, is used to hit
a car travelling along the same road away from the
gun with a horizontal speed of 72 kmh −1 . The car is
at a distance of 500 m from the gun when the gun is
fired at an angle of 45° with the horizontal. Find the
distance of the car from the gun when the shell hits
it and the speed of the projection of the shell. (Take
and
2 = 1.41 )
For the shell to hit the car, time taken by the shell to
fly from O → A → B must equal the time taken by
the car to go from C to B with a uniform speed of
20 ms −1 . So,
tcar from = tshell from = T =
C →B
05_Kinematics 2_Part 1.indd 22
O→ A→ B
OC + CB = Range ( R )
⇒
500 + ( 20 ) T =
u2 sin ( 90° )
g
2
20 2u u
=
g
g
⇒
500 +
⇒
2
u − 20 2u − 500 g = 0 ⇒
u2 − 20 2u − 4900 = 0 ⇒
u=
20 2 + 800 + 4 ( 4900 )
2
⇒
u=
20 2 + 142.8 28.2 + 142.8
=
= 85.5 ms −1
2
2
When the shell hits the car, distance of car from the
gun is equal to the range
R=
u2
= 746 m
g
Illustration 15
The velocity of a projectile when it is at the greatest
2
times its velocity when it is at half of its
5
greatest height. Determine its angle of projection.
height is
Solution
x
Also, we observe that when the shell hits the car at
B , then
Substituting in (1), we get
g = 9.8 ms
B
500 m
v = 20 ms −1 −2
C
2u
2u sin ( 45° )
=
g
g
Solution
Let the particle be projected with velocity u at an
angle θ with the horizontal. Horizontal component
of its velocity at all heights will be u cos θ .
At the greatest height, the vertical component of
velocity is zero, so the resultant velocity is
11/28/2019 7:17:00 PM
Chapter 5: Kinematics II
v = u cos θ
1
At half the greatest height during upward motion, we
have
5.23
x
30°
10 ms–1
4 ms–1
y
vy = uy2 − 2 ay y
⇒
⇒
vy = u2 sin 2 θ − 2 ( g )
vy =
u2 sin 2 θ
4g
u sin θ
2 Hence, resultant velocity at half of the greatest height
is
u2 sin 2 θ
u cos θ +
2
vx2 + vy2 =
v2 =
Since, it is given that
⇒
⇒
v12
2
2
v1
=
v2
2
5
u2 cos 2 θ
2
=
=
2
2
2
v2
u sin θ 5
u2 cos 2 θ +
2
1
1
1 + tan 2 θ
2
=
⇒
2 + tan 2 θ = 5 ⇒
tan 2 θ = 3 2
5
1 2
ay t
2
1
⇒ 10 = 2t + ( 9.8 ) ( t 2 )
2
Solving these equations we get
t = 1.24 s
and x = ux t = 4.29 m Illustration 17
The range of the rifle bullet is 1000 m , when θ is the
angle of projection. If the bullet is fired with the same
angle from a car travelling horizontally at a speed of
36 kmh −1 towards the target, calculate the amount
by which the range is increased.
Solution
R = 1000 =
Illustration 16
A person working on the roof of a house suddenly
drops his hammer, which slides down the roof at a
constant speed of 4 ms −1 . The roof makes an angle of
30° with the horizontal, and its lowest point is 10 m
above the ground. What is the horizontal distance
travelled by the hammer after it leaves the roof of the
house before it hits the ground?
Solution
ux = 4 cos 30° = 2 3 ms
−1
uy = 4 sin 30° = 2 ms
2
( u cos θ )( u sin θ ) …(1)
g
When the bullet is fired from a car travelling at a
speed of 36 kmh −1 ( = 10 ms −1 ) at the same angle,
then the new horizontal component of the velocity of
the bullet becomes
⇒ tanθ = 3 ⇒ θ = 60° 05_Kinematics 2_Part 1.indd 23
Since y = uy +
−1
cos θ + 10
ux′ = u
However, the y component remains unaltered. So, if
R ′ is the new range, then
R′ =
2
( u cos θ + 10 )( u sin θ )
g
⇒
R′ =
2
20
( u cos θ )( u sin θ ) + u sin θ
g
g
⇒
R′ − R =
20
u sin θ …(2)
g
11/28/2019 7:17:06 PM
5.24 JEE Advanced Physics: Mechanics – I
From (1),
⇒
500 g
sin θ cos θ
u=
⇒
500 g
20
sin θ
g sin θ cos θ
⇒
⇒
R′ − R =
⇒
500
5000
( tan θ ) = 20
R ′ − R = 20
tan θ
g
98
⇒
R ′ − R = 20
⇒
1000
tan θ m
R′ − R =
7
v = u 1 + cosec 2 α − 2 v = u cosec 2 α − 1 v = u cot α ∵ 1 + cot 2 α = cosec2 α
{
}
Illustration 19
With what minimum speed must a particle be projected from origin so that it is able to pass through a
given point P ( x , y ) ?
2500
tan θ
49
Solution
Let u and θ be the velocity and angle of projection
respectively.
For the projectile to pass through P ( x , y )
Illustration 18
It at any point on the path of a projectile its velocity be u and inclination be α . Show that the particle will move at right angles to the former direction
u
after a time t =
when its velocity would be
g sin α
v = u cot α .
Solution
y = x tan θ −
⇒
2u2
( 1 + tan 2 θ )
(
)
gx 2 tan 2 θ − 2xu2 tan θ + gx 2 + 2 yu2 = 0
The projectile will pass through P(x, y) if this equation (quadratic in tanq) gives some real value of θ ,
i.e., its discriminant ≥ 0.
(
The initial velocity of the particle is
iˆ u = ( u cos α ) iˆ + ( u sin α ) ˆj …(1)
At any time t , the velocity of the particle is v . Then
v = vx iˆ + vy ˆj
⇒ iˆ v = ( u cos α ) iˆ + ( u sin α − gt ) ˆj …(2)
Now, v ⊥ u , so we have
v⋅u = 0
gx 2
)
4 x 2 u 4 − 4 gx 2 gx 2 + 2 yu2 ≥ 0
u 4 − 2 gyu2 − g 2 x 2 ≥ 0
u 4 − 2 gyu2 + g 2 y 2 ≥ g 2 x 2 + g 2 y 2
( u2 − gy ) ≥ ( x 2 + y 2 ) g 2
2
⇒
u ≥ gy + g x 2 + y 2
⇒
u2 cos 2 α + u2 sin 2 α − ugt sin α = 0 So, the minimum value of u is uMIN = gy + g x 2 + y 2 .
⇒
u2 − ugt sin α = 0 Illustration 20
⇒
t=
⇒
v = u2 + g 2 t 2 − 2ugt sin α
Two shots are projected from a gun at the top of a hill
with the same velocity u at angles of projection α
and b respectively. If the shots strike the horizontal
ground through the foot of the hill at the same point,
show that the height h of the hill above the plane is
given by
u
g sin α
Further, from (2), we have
v = u2 cos 2 α + u2 sin 2 α + g 2 t 2 − 2ugt sin α
⇒
v = u2 + g 2
05_Kinematics 2_Part 1.indd 24
h=
u2
⎛ u ⎞
− 2ug ⎜
sin α
2
2
⎝ g sin α ⎟⎠
g sin α
2u2 ( 1 − tan α tan β )
g ( tan α + tan β )
2
11/28/2019 7:17:14 PM
Chapter 5: Kinematics II
Solution
Let R be the horizontal range for both. Taking the
point of projection as origin, the point struck is
( R, h ) . This point satisfies the equations of both the
trajectories.
− h = R tan α −
1
R2
g 2
…(1)
2 u cos 2 α
and − h = R tan β −
1
R2
g 2
…(2)
2 u cos 2 β
How must he aim his gun so that both the shots hit
the bird simultaneously? What is the distance of the
foot of the tower from the two boys and the height of
the tower? With what velocities and when do the two
shots hit the bird?
Solution
The situation is shown in figure
y
Bird
From these equations, we have
R tan α −
⇒
⇒
1
R2
1
R2
g 2
= R tan β − g 2
2
2 u cos α
2 u cos 2 β
1
1
R2
R2
R ( tan α − tan β ) = g 2
−
g
2 u cos 2 α 2 u2 cos 2 β
100 ms–1
O
2
2u
( tan α − tan β ) = R ( sec2 α − sec2 β )
g
⇒
2u2
= R ( tan α + tan β ) …(3)
g
⇒
2u2
R=
g ( tan α + tan β )
)
Substituting the value of R in equation (1), we get
h=
⇒
h=
⇒
h=
g ( tan α + tan β )
2
(
sec 2 α − tan 2 α − tan α tan β
For first shot,
horizontal displacement = x1
vertical displacement = h
For second shot,
horizontal displacement = x2
vertical displacement = h
For first shot,
)
2
h = ( 100 sin 30° ) t −
( 1 − tan α tan β )
Illustration 21
Two boys simultaneously aim their guns at a bird sitting on a tower. The first boy releases his shot with a
speed of 100 ms −1 at an angle of projection of 30° .
The second boy is ahead of the first by a distance of
50 m and releases his shot with a speed of 80 ms −1 .
05_Kinematics 2_Part 1.indd 25
x2
1 2
gt …(2)
2
For second shot,
2u2
g ( tan α + tan β )
x
The two shots will hit the bird simultaneously at a
particular instant t , if the horizontal and vertical displacements of the two shots are as shown.
⎛
⎞
sec 2 α
2u 2
− tan α ⎟
⎜
⎠
g ( tan α + tan β ) ⎝ tan α + tan β
2u 2
θ
2
Here, x1 = x2 + 50 …(1)
gR
⎛
⎞
h = R⎜ 2
− tan α ⎟
2
⎝ 2u cos α
⎠
⇒
30°
1
h
x1
2u2
⋅ ( tan α − tan β ) = R ⎡⎣ 1 + tan 2 α − 1 + tan 2 β ⎤⎦
g
) (
80 ms–1
50 m
⇒
(
5.25
h = ( 80 sin θ ) t −
1 2
gt …(3)
2
1
2
1
2
⇒
( 100 sin 30° ) t − gt 2 = ( 80 sin θ ) t − gt 2
⇒
100 sin 30° 50
sin θ =
=
= 0.625
80
80
⇒ θ = sin −1 ( 0.625 ) = 38°68 ′ 11/28/2019 7:17:20 PM
5.26 JEE Advanced Physics: Mechanics – I
Now, x1 = ( 100 cos 30° ) t and x2 = 80 cos ( 38°68 ′ ) t
Method II
d2 y
Since x1 = x2 + 50
⇒
( 100 cos 30° ) t = 50 + 80 cos ( 38°68 ′ ) t
⇒
⎤
⎡
⎛ 3⎞
t ⎢ 100 ⎜
⎟⎠ − 80 × 0.7806 ⎥ = 50
⎝
2
⎣
⎦
⇒
t [ 86.6025 − 62.4499 ] = 50 ⇒
t=
dt
⇒
⇒
50
= 2.07 s
24.1526
⇒
Illustration 22
A particle moves in the plane x-y with constant acceleration a directed along negative y-axis. The equation of motion of particle has the form y = Ax − Bx 2 .
A and B are positive constants. Find the velocity of
the particle at the origin of the co-ordinates.
Solution
Method I
Comparing the above given equation with the equation of the oblique projectile moving under the influ
ence of the acceleration field a.
2u cos 2 θ {Equation for Oblique Projectile}
a
⇒
A = tan θ and B =
⇒
A = tan θ and sec 2 θ =
2
2u cos θ 2u2 B
…(1)
a
sec 2 θ − tan 2 θ = 1
2u B
− A2 = 1 a
⇒
(1 + A )
u=
05_Kinematics 2_Part 1.indd 26
2
dy
dx
dx
=A
− 2Bx
…(2)
dt
dt
dt
d2 y
dt
=A
2
a
2B ⎡ ⎛ dx ⎞
d2 x ⎤
−
2
+
B
x
⎢
⎥
⎜
⎟
dt 2
dt 2 ⎥⎦
⎢⎣ ⎝ dt ⎠
2
d2 x
⎤
⎡ ⎛ dx ⎞ 2
− a = 0 − 2B ⎢ ⎜
⎟⎠ + 0 ⎥ {from given relations}
⎝
⎢⎣ dt
⎥⎦
⇒
ux2 =
a
…(3)
2B
Also from (2),
⇒
dy
a
= uy = Aux = A
2B
dt x = 0
⇒
uy = A
{at Origin}
a
{∵ at Origin x = 0 }…(4)
2B
Now, u2 = ux2 + uy2
⇒
u2 =
a
a
+ A2
2B
2B
⇒
u=
( 1 + A ) 2aB
{using (3) & (4)}
2
RANGE, MAXIMUM HEIGHT AND TIME OF
FLIGHT FOR COMPLIMENTARY ANGLES
2
We know that
⇒
= 0 {Given}
y = Ax − Bx 2 …(1)
ax 2
2
2
dt 2
⇒
h = ( 103.5 − 20.996 ) = 82.504 m y = x tan θ −
d2 x
Again differentiating both sides of equation (2) w.r.t. t,
x2 = 179.27 − 50 = 129.27 m
1
2
Also, h = 100 sin 30°t − × 9.8 × ( 2.07 )
2
⇒
= −a ;
Differentiating both sides of equation (1) w.r.t. t,
So, x1 = 100 cos 30° × 2.07 = 179.27 m and
⇒
2
For complimentary angles, say ϕ and ( 90 − ϕ ) , let the
corresponding ranges, maximum heights and times
of flight be Rϕ , R90−ϕ , Hϕ , H90−ϕ and Tϕ , T90−ϕ , then
Rϕ =
{using (1)}
⇒
2u2 cos ϕ sin ϕ
2u2 sin ϕ cos ϕ
and R90 −ϕ =
g
g
Rϕ = R90 −ϕ =
2u2 sin ϕ cos ϕ
…(1)
g
11/28/2019 7:17:30 PM
Chapter 5: Kinematics II
So, we must remember that for complimentary angles
range is the same i.e.,
R1° = R89° , R15° = R75° and so on
Further more,
u2 sin 2 ϕ
…(2)
2g
Hϕ =
u2 sin 2 ( 90 − ϕ ) u2 cos 2 ϕ
…(3)
H90 −ϕ =
=
2g
2g
From (2) and (3), we observe that
u2
…(4)
Hϕ + H90 −ϕ =
2g
Also, from equations (1), (2) and (3), we observe that
Rϕ = R90 −ϕ = 4 Hϕ H90 −ϕ …(5)
⇒
R1° = R89° = 4 H1° H89°
⇒
⇒
2u cos ϕ
2u sin ϕ
and T90 −ϕ =
g
g
Tϕ T90 −ϕ =
Tϕ T90 −ϕ =
i.e., T1° T89° =
4u2 sin ϕ cos ϕ
g2
2Rϕ
g
=
2R90 −ϕ
g
=
2 ⎛ 2u2 sin ϕ cos ϕ ⎞
⎟⎠
g ⎜⎝
g
So, for complimentary angles f and 90 − ϕ , following
points are worthnoting.
u2
2g
(c) Rϕ = R90 −ϕ = 4 Hϕ H90 −ϕ
(d) Tϕ T90 −ϕ =
05_Kinematics 2_Part 1.indd 27
2Rϕ
g
=
2R90 −ϕ
u2 sin ( 2θ )
g
Since R =
( 39.2 )2 sin ( 2θ )
⇒
78.4 =
⇒
sin ( 2θ ) =
⇒
9.8
1
2
2θ = 30° or 150° ⇒ θ = 15° or 75° t=
Problem Solving Technique(s)
(b) Hϕ + H90 −ϕ =
Solution
⇒
…(6)
2u2 sinϕ cos ϕ
g
A shell bursts on contact with the ground and the
fragments fly in all the directions with speeds upto
39.2 ms −1 . Show that a man 78.4 m away is in danger for 4 2 s .
t = t75° − t15°
2R1° 2R89°
=
g
g
(a) Rϕ = R90 −ϕ =
Illustration 23
i.e., the range will be same for these two complimentary angles of 15° and 75° . However, the time of
flight will not be the same. Hence the person 78.4 m
away will be in danger for a time t given by
Finally,
Tϕ =
5.27
2u
( sin ( 75° ) − sin ( 15° ) )
g
2 ( 39.2 ) ⎡
⎛ 75° + 15° ⎞
⎛ 75° − 15° ⎞ ⎤
⎟⎠ sin ⎜⎝
⎟⎠ ⎥
⎢ 2 sin ⎜⎝
2
2
⎦
⎣
⇒
t=
⇒
t = 16 sin ( 45° ) sin ( 30° ) ⇒
t=
9.8
8
2
=
8 2
=4 2s
2
TWO UNIQUE TIMES FOR WHICH
PROJECTILE IS AT SAME HEIGHT
It has been observed that, there are two unique times
at which the projectile is at same height h ( < H ) . Let
us consider a projectile launched with initial velocity
u , making an angle θ with the horizontal. Let the
projectile be at a height h ( < H ) at time t. Then we
observe just by a qualitative argument that this is true
for upward motion of the projectile and then for the
downward motion, as shown.
g
11/28/2019 7:17:38 PM
5.28 JEE Advanced Physics: Mechanics – I
The following Illustration shows how to use
­equations (2), (3), (4), (5) and (6) to get the results
efficiently.
y
A
B
Illustration 24
u
h
O
h
x
θ
x
t1
t2
Mathematically, at time t, we have
1 2
gt
2
2h
⎛ 2u sin θ ⎞
t2 − ⎜
t+
= 0 …(1)
⎟
g
g
⎝
⎠
h = ( u sin θ ) t −
⇒
This equation happens to be a quadratic equation in t
(with non-zero discriminant), hence has two unique
roots t1 and t2 . A very important observation which
cannot be skipped here is that
t1 + t2 =
t1t2 =
2u sin θ
= T …(2)
g
A stone is projected from a point on the ground in
such a direction so as to hit a bird on the top of a telegraph post of height h and then attain a maximum
height of 2h above the ground. If at the instant of
projection the bird were to fly away horizontally with
a uniform speed then calculate the ratio between the
horizontal velocity of the bird and the stone, if the
stone still hits the bird while descending.
Solution
H = 2h =
⇒
u2 sin 2 θ
2g
u sin θ = 2 gh …(1)
Since h = ( u sin θ ) t +
⇒
h = 2 ght −
2h
…(3)
g
1 2
gt
2
y
From the diagram we must understand that,
t1 is the time taken by the projectile to fly from
O → A and
t2 is the time taken by the projectile to fly from
O→A→B
So, time taken by the projectile to fly from A to
B is
A
u
O
θ
t1
2h
B
C
h
h
M
N
2
t = t2 − t1 =
2
4u2 sin 2 θ
2
−
⇒
8h
…(5)
g
g
If, x be the horizontal separation between the two
events, then
x = ( u cos θ ) t = ( u cos θ )
05_Kinematics 2_Part 1.indd 28
4u2 sin 2 θ
g2
−
8h
…(6)
g
t2 − 4
D
x
t2
t = t2 − t1 …(4)
Since ( t2 − t1 ) = ( t2 + t1 ) − 4t1t2
1
( − g ) t2
2
h 2h
t+
= 0 …(2)
g
g
This quadratic equation in t will have two roots, t1
and t2 . So,
4
t1 =
h
16 h 8 h
−
−
g
g
g
2
= (2 − 2 )
h
…(3)
g
11/28/2019 7:17:43 PM
5.29
Chapter 5: Kinematics II
h
16 h 8 h
+
−
g
g
g
4
t2 =
2
At Any Position at Vertical Height h(< H)
= (2+ 2 )
h
…(4)
g
Here too everything remains the same i.e.,
r=
So, time taken by the stone to fly from B to C is (t2 - t1).
⇒
BC = ( u cos θ ) ( t2 − t1 ) …(5)
After reading the question carefully, we observe that
at the instant of projection of stone the bird were to
fly away horizontally with a uniform speed (say vb )
and the stone, during its downward motion (at C)
still hits the bird. So, time taken by the bird to fly
from B to C is the time taken by the stone to go from
O → B → A → C . Hence, for the bird, we have
vT2 v⊥2
v2
=
=
aC
a
g cos β
But here, v 2 = u2 − 2 gh At Maximum Height (H )
At the maximum height H , velocity is u cos θ .
⇒
vT = v⊥ = u cos θ y
BC = vb t2 …(6)
g
From (5) and (6), we get
P
vb t2 = ( u cos θ ) ( t2 − t1 )
⇒
vb
t −t
2 2
= 2 1 =
=
u cos θ
t2
2+ 2
u
Consider a projectile launched with initial velocity u,
making an angle q with the horizontal. Let the projectile be at a point P, at any instant t (say). Then we
know that
vT2 v⊥2
v2
=
=
…(1)
aC
a
g cos β
where
v 2 = ( u sin θ − gt ) + ( u cos θ ) {studied earlier}
2
y
g
v
P
u
β
r β
⇒
θ
O
θ
x
Also, aC = a = g
⇒
r=
vT2 v⊥2 u2 cos 2 θ
=
=
aC
a
g
This result can also be obtained from cases discussed
in (A) and (B), where we can put special values to get
the desired result. (I hope this is a small exercise for
you people).
Illustration 25
A particle is projected with a speed u making an
angle θ with the horizontal. Find the radius of curvature at the point where the particle velocity makes
θ
an angle
with the horizontal. Assume the parti2
cle to move under gravity in the absence of any air
drag.
Solution
g g cos β
O
vT = v⊥ = u cos θ
C
2 +1 At Any Instant of Time (t)
2
ac = aıı = g
2
Radius of Curvature of an Oblique
Projectile at a Point P
r=
{studied earlier}
x
First of all, after reading the question carefully, we see
that the particle will act as an oblique projectile following a parabolic path, as shown.
v 2 = u2 + g 2 t 2 − 2ugt sin θ 05_Kinematics 2_Part 1.indd 29
11/28/2019 7:17:48 PM
5.30 JEE Advanced Physics: Mechanics – I
y
Since R =
g
P
u
O
ac
r=
⇒
ac = g cos θ
2
g
x
vT2 v⊥2
=
ac
a
v2
…(1)
⎛θ⎞
g cos ⎜ ⎟
⎝ 2⎠
{∵ Horizontal Motion is Non-Accelerated}
v=
u cos θ
…(2)
⎛θ⎞
cos ⎜ ⎟
⎝ 2⎠
Substituting (2) in (1), we get
r=
2
A ball is thrown from ground level so as to just clear
a wall 4 m high at a distance of 4 m and falls at a
distance of 14 m from the wall. Find the magnitude
and direction of the velocity of the ball.
Solution
⎛θ⎞
Since v cos ⎜ ⎟ = u cos θ
⎝ 2⎠
⇒
x⎞
⎛
y = x tan θ ⎜ 1 − ⎟
⎝
R⎠
Illustration 26
θ
Since r =
⇒
θ
2
vT = v
θ /2
2u2 sin θ cos θ
g
2
u cos θ
⎛θ⎞
g cos 3 ⎜ ⎟
⎝ 2⎠
As the ball strikes the ground at a distance 14 m from
the wall, the range is R = ( 4 + 14 ) m = 18 m.
x⎞
⎛
Since y = x tan θ ⎜ 1 − ⎟ …(1)
⎝
R⎠
where, x = 4 m , y = 4 m and R = 18 m.
⇒
4 ⎞
⎛
⎛ 7⎞
4 = 4 tan θ ⎜ 1 − ⎟ = 4 tan θ ⎜ ⎟
⎝
⎠
⎝ 9⎠
18
⇒
9
tan θ =
7
⇒
sin θ =
Again R =
9
130
and cos θ =
7
130 2 2
u sin θ cos θ
g
2
9
7
× u2 ×
×
9.8
130
130 Equation of Trajectory of an
Oblique Projectile in Terms of Range
⇒
R=
Since, we know that for a projectile launched with
­initial velocity u , making an angle θ with the
­horizontal, the equation of trajectory is given by
⇒
u2 =
18 × 9.8 × 130 × 130
2×9×7
⇒
u2 =
98 × 13
= 182
7
⇒
u = 182 ms −1 = 13.5 ms −1 y = x tan θ −
⇒
⇒
gx 2
2u2 cos 2 θ
⎤
⎡
gx 2
y = x tan θ ⎢ 1 − 2
⎥
2 (
2u cos θ x tan θ ) ⎦ ⎣
x
⎤
⎡
y = x tan θ ⎢ 1 −
2
⎛ 2u sin θ cos θ ⎞ ⎥
⎢
⎥
⎜⎝
⎟⎠ ⎥
⎢⎣
g
⎦
05_Kinematics 2_Part 1.indd 30
Illustration 27
A particle is thrown over a triangle from one end of
a horizontal base and after grazing the vertex falls on
the other end of the base. If α and β be the base
angles and θ the angle of projection, prove that
tan θ = tan α + tan β .
11/28/2019 7:17:55 PM
Chapter 5: Kinematics II
Solution
The situation is shown in figure. In triangle NOP ,
y
tan α = .
x
P(x, y)
R = Range
θ
y
α
β
O
B
N
x
x
R–x
x
y
R−x
y
y
tan α + tan β = +
x R−x yR
…(1)
x(R − x)
x⎞
⎛
Since, the equation of trajectory is y = x tan θ ⎜ 1 − ⎟
⎝
R⎠
yR
…(2)
⇒ tan θ =
x(R − x)
tan α + tan β =
From equations (1) and (2), we get
ax = 0
ay = 0
dvx
=0
dt
vx = constant = k1
dvy
(say)
⇒
dx
= k1
dt
05_Kinematics 2_Part 1.indd 31
⇒
⇒
dt
∫
t
∫
dy = k 2 dt
0
0
0
y = k 2 t …(2)
⇒
y k2
=
= k (say)
x k1
⇒
y = kx , equation of a straight line with slope
k2
k1
Method II
Let us now discuss the relative motion between two
projectiles or the path observed by one projectile of
the other. Suppose that two particles are projected
from the ground with speeds u1 and u2 at angles α 1
and α 2 as shown in figure. Acceleration of both the
particles is g downwards. So, relative acceleration
between them is zero because
a12 = a1 − a2 = g − g = zero
y
u2
u1
α1
α2
x
i.e., the relative motion between the two particles is
uniform. Now,
( u1 )x = u1 cos α1 and ( u2 )x = u2 cos α 2
( u1 )y = u1 sin α1 and ( u2 )y = u2 sin α 2
a12 = 0
=0
vy = constant = k 2
dy
= k2
dt
x
y
(say)
⇒
⇒
y
Method I
Since both the projectiles are moving under the influence of gravity, so acceleration of one projectile as
seen from the other is
ˆ + a ˆj = 0
a
=
g
−
g
=
0
⇒
a
i
x
y
⇒
∫
dx = k1 dt
x = k1t …(1) ⇒
Relative Motion Between Two
Projectiles/Motion of One Projectile
as Seen from Another Projectile
⇒
y
t
⇒
tan θ = tan α + tan β
⇒
∫
⇒
dy = k 2 dt
So, motion of one projectile as seen from another projectile is a straight line passing through the origin (or
the point of launch).
In triangle BPN , tan β =
⇒
⇒
dx = k1 dt
0
y
⇒
⇒
5.31
(u12)y
u12
θ
(u12)x
x
11/28/2019 7:18:04 PM
5.32 JEE Advanced Physics: Mechanics – I
Therefore,
B
( u12 )x = ( u1 )x − ( u2 )x = u1 cos α1 − u2 cos α 2
and ( u12 )y = ( u1 )y − ( u2 )y = u1 sin α 1 − u2 sin α 2
( u12 )x and ( u12 )y are the x and y components of
relative velocity of 1 with respect to 2.
Hence, relative motion of 1 with respect to 2 is
⎡ ( u12 )y ⎤
⎥ with
a straight line at an angle θ = tan −1 ⎢
⎢⎣ ( u12 )x ⎥⎦
positive x-axis.
(a) If ( u12 )x = 0 , i.e., u1 cos α 1 = u2 cos α 2 , then the
relative motion is along y-axis or in vertical direction (as θ = 90° ).
(b) Similarly, if ( u12 )y = 0 , i.e., u1 sin α 1 = u2 sin α 2 ,
then the relative motion is along x-axis or in horizontal direction (as θ = 0° ).
CONDITION OF COLLISION between
TWO PROJECTILES
Now it is clear that relative motion between two
projectiles is uniform and the path of one projectile
as observed by the other is a straight line. Let the
particles be projected simultaneously from two different heights h1 and h2 with speeds u1 and u2 in
the directions shown in figure. Then the particles collide in air if the relative velocity of 1 with respect to
2 ( u12 ) is along line AB or the relative velocity of 2
with respect to 1 ( u21 ) is along the line BA . Thus,
tan θ =
( u12 )y ⎛ h2 − h1 ⎞
=⎜
⎟
( u12 )x ⎝ s ⎠
α1
h2
h1
s
05_Kinematics 2_Part 1.indd 32
h2 – h1
x
θ
u12x
s
Here, ( u12 )y = u1 sin α 1 − u2 sin α 2 and
⇒
( u12 )x = ( u1 cos α1 ) − ( −u2 cos α 2 )
( u12 )x = u1 cos α1 + u2 cos α 2
If both the particles are initially at the same level
( h1 = h2 ) , then for collision
( u12 )y = 0 or u1 sin α1 = u2 sin α 2
The time of collision of the two particles will be
AB
t= =
u12
AB
( u12 )x + ( u12 )y
Further, the above conditions are not merely sufficient
for collision to takes place. For example, the time of
collision discussed above should be less than the time
of collision of either of the particles with the ground.
2
2
Illustration 28
Two guns situated at the top of a hill of height 10 m,
(a) the time interval between the firings and
(b) the coordinates of the point P. Take origin of
coordinates system at the foot of the hill right
below the muzzle and trajectory in x-y plane.
B
u1
A
A
u12
fire one shot each with the same speed 5 3 ms −1 at
some interval of time. One gun fires horizontally and
the other fires upwards at an angle of 60° with the
horizontal. The shots collide in air at a point P. Find
u2
α2
y
u12y
⇒
Solution
Let the shot fired at 60° above horizontal reach point
P in a time t . Then the shot fired horizontal must
have taken a time ( t − Δt ) to reach the point P, where
Δt is the time log between firing of shots.
11/28/2019 7:18:10 PM
5.33
Chapter 5: Kinematics II
(a) Let they collide at the point P ( x , y ) , then
A
x = (5 3 cos 60) t = 5 3 (t − Δt)
⇒ t = 2t − 2Δt ⇒ t = 2Δt …(1)
1 2
⇒ y = −5 3 sin 60 t + 2 gt
1
2
⇒ y = 2 g(t − Δt)
(
22 m
15
2
2
⇒ − 2 t + 5t = 5(t − Δt)
2
2
⇒ −6t + 4t = t 2
⇒ 3t − 6t = 0
{∵ t ≠ 0 }
So, the interval between the firings is
(b) x = 5 3 cos 60 t = 5 3 m
2
( uBA )H = uB cos 45° − ( −uA cos 45° )
)
)
Illustration 29
Two particles are simultaneously thrown from the
roofs of two high buildings as shown in figure. Their
velocities are vA = 2 ms −1 and vB = 14 ms −1 respectively. Calculate the minimum distance between the
particles in the process of their motion. Also find the
time when they are at closest distance.
2
= 8 2 ms −1
x = 22 − ( uBA )H t = ( 22 − 8 2t ) m
and vertical distance between A and B after time
t is
(
)
y = 9 − ( uBA )V t = ( 9 − 6 2t ) m
Therefore, distance between them after time t is
(
(
1
Horizontal distance between A and B after time t
Hence the particles collide at 5 3 , 5 m.
05_Kinematics 2_Part 1.indd 33
1
−1
( uBA )H = ( 14 + 2 )
⇒ 3t(t − 2) = 0
⇒ t = 2 s t
=1s
2
( uBA )V = uB sin 45° − uA sin 45° = ( 14 − 2 )
( uBA )V = 6 2 ms
Relative velocity of B with respect to A in horizontal direction,
3
t2
2
⇒
−
t
+
t
=
2
4 ⇒ y = 5 m 11 m
Assuming A to be at rest
aBA = aB − aA = 0 as aA = aB = g (downwards)
Thus, the relative motion between them is uniform.
Relative velocity of B with respect to A in vertical
direction,
P(x, y)
(
B
Solution
60°
5√3 ms–1
Δt =
45°
20 m
)
5√3 ms–1
45°
)
r = x2 + y2
⇒
r 2 = x 2 + y 2 = ( 22 − 8 2t ) + ( 9 − 6 2t ) 2
For r to be minimum
2
d ( 2)
r =0
dt
⇒
2 ( 22 − 8 2t ) ( −8 2 ) + 2 ( 9 − 6 2t ) ( −6 2 ) = 0 ⇒
88 − 32 2t + 27 − 18 2t = 0 11/28/2019 7:18:19 PM
5.34 JEE Advanced Physics: Mechanics – I
23
⇒
t=
⇒
rmin = x 2 + y 2 at time t =
10 2
s
⇒
23
10 2
Substituting the values, we get
s
Again y1 =
From points A and B , at the respective heights of
2 m and 6 m , two bodies are thrown simultaneously towards each other; one is thrown horizontally
with a velocity of 8 ms −1 and the other, downward at
an angle of 45° to the horizontal at an initial velocity such that the bodies collide in flight. The horizontal distance between points A and B equals 8 m .
Calculate the initial velocity v0 of the body thrown
at an angle 45° , the co-ordinates ( x , y ) of the point
of collision, the time of flight t of the bodies before
colliding and velocities vA , vB of the two bodies at
the instant of collision. The trajectories lie in a single
plane.
Solution
⇒
From equations (2) and (3), we get
2−
⇒
1 2
1
⎛ v ⎞
gt = 6 − ⎜ 0 ⎟ t − gt 2
⎝ 2⎠
2
2
⎛ v ⎞
4 = ⎜ 0 ⎟ t …(4)
⎝ 2⎠
Substituting this value of t in equation (1), we get
8 = 8t + 4
⇒
t = 0.5 s v0
8 ms–1
⎛ v ⎞
4 = ⎜ 0 ⎟ × 0.5
⎝ 2⎠
B
v0
2
⇒
45°
v0
2 y2
v0 = 11.28 ms −1 Now, x = 8t = 8 × 0.5 = 4 m
6m
y1
P
x
(x, y)
y = 2−
x1
⇒
(0, 0)
8m
Let the two bodies collide after t second.
⎛ v ⎞
From figure, x = 8 × t and x1 = ⎜ 0 ⎟ × t
⎝ 2⎠
⎛ v ⎞
x + x1 = 8t + ⎜ 0 ⎟ t
⎝ 2⎠
But given that x + x1 = 8
05_Kinematics 2_Part 1.indd 34
1
⎛ v ⎞
y = 6 − ⎜ 0 ⎟ t − gt 2 …(3)
⎝ 2⎠
2
From equation (4),
The situation is shown here.
⇒
1 2
gt …(2)
2
1
⎛ v ⎞
Further, y 2 = ⎜ 0 ⎟ t + gt 2
⎝ 2⎠
2
Illustration 30
2m
1 2
gt
2
and y = 2 − y1 = 2 −
rmin = 6 m
A
⎛ v ⎞
8 = 8t + ⎜ 0 ⎟ t …(1)
⎝ 2⎠
1 2
gt = 0.775 m
2
( x , y ) = ( 4, 0.775 ) 2
2
At P , we have vA = vAx
+ vAy
⇒
2
2
vA = ( 8 ) + ( 4.9 ) = 9.4 ms −1 2
2
and vB = vBx
+ vBy
⇒
⇒
2
⎛ 11.28 9.8 ⎞
⎛ 11.28 ⎞
vB = ⎜
+⎜
+
⎟
⎝ 2
⎝ 2 ⎠
2 ⎟⎠
2
−1
vB = 15.2 ms 11/28/2019 7:18:28 PM
Chapter 5: Kinematics II
5.35
Test Your Concepts-III
Based on Oblique Projectile
1. Find the angle of projection of a projectile for
which the horizontal range and maximum height
are equal.
2. Prove that the maximum horizontal range is four
times the maximum height attained by the projectile.
3. Two projectiles are launched with same initial
velocity at different launch angles so as to have
identical range. Show that the sum of the maximum heights attained by them is independent of
the launch angles.
4. Show that there are two values of time for which
a projectile is at the same height. Also prove that
the sum of these two times is equal to the time of
flight. Also find the time lapse and the horizontal
separation between the two events.
5. Two shots are fired simultaneously from the top
and bottom of a vertical tower AB at angles a and
b with horizontal respectively. Both shots strike at
the same point C on the ground at distance s from
the foot of the tower at the same time. Show that
the height of the tower is S ( tan β − tanα ) .
6. A ball is shot from the ground into the air. At a
height of 9.1 m, its velocity is observed to be
iˆ v = ( 7.6iˆ + 6.1ˆj ) ms −1.
(a) To what maximum height does the ball rise?
(b) What total horizontal distance does the ball
travel?
(c) Find the magnitude and direction of the ball’s
velocity just before it hits the ground?
7. The coach throws a baseball to a player with an initial speed of 20 ms −1 at an angle of 45° with the
horizontal. At the moment the ball is thrown, the
player is 50 m from the coach. At what speed and
in what direction must the player run to catch the
ball at the same height at which it was released?
8. A projectile is fired from the vertical tube mounted
on the vehicle which is travelling at the constant
speed u = 30 kmh−1. The projectile leaves the tube
with a velocity vr = 20 ms −1 relative to the tube. If
air resistance is neglected, show that the projectile
will land on the vehicle at the tube location and
05_Kinematics 2_Part 1.indd 35
(Solutions on page H.146)
calculate the distance s travelled by the vehicle
during the flight of the projectile.
Vr
u
9. A stone is thrown from the top of a tower of height
50 m with a velocity of 30 ms −1 at an angle of 30°
above the horizontal. Find the
(a) time during which the stone will be in air,
(b) distance from the tower base to where the
stone will hit the ground,
(c) speed with which the stone will hit the ground.
( Take g = 10 ms −2 ).
10. At the same instant two boys throw balls A and B
from the window with a speed v0 and kv0, respectively, where k is a constant. Show that the balls
cosθ
with collide if k =
.
cos ϕ
A
θ
v0
h
kv0
ϕ
B
d
11. A projectile aimed at a mark which is in the horizontal plane through the point of projection falls
a cm short of it when the elevation is a and goes
b cm far when the elevation is b. Assume the
speed of projection to be the same in all the cases,
then show that the proper angle of projection is
1 −1 ⎛ a sin( 2β ) + b sin( 2α ) ⎞
sin ⎜
⎟⎠ .
⎝
2
a+b
11/28/2019 7:18:30 PM
5.36 JEE Advanced Physics: Mechanics – I
12. Two particles are simultaneously projected in the
same vertical plane from the same point with
velocities u and v at angles a and b with horizontal.
Find the time that elapses when their velocities are
parallel.
13. A projectile takes off with an initial velocity of
10 ms −1 at an angle of elevation of 45°. It is just
able to clear two hurdles of height 2 m each, separated from each other by a distance d. Calculate d.
At what distance from the point of projection is the
first hurdle placed? Take g = 10 ms −2 .
14. The acceleration of gravity can be measured by
projecting a body upward and measuring the time
it takes to pass two given points in both d
­ irections.
Show that if the time the body takes to pass a
horizontal line A in both directions is tA and time
to go by a second line B in both directions is tB,
then assuming that the acceleration is constant, its
8h
­magnitude is g = 2 2 , where h is the height of
t A − tB
the line B above line A.
Height
B
tB
A
h
tA
Time
15. To meet design criteria, small ball bearings must
bounce through an opening of limited size at the
top of their trajectory when rebounding from a
heavy plate as shown. Calculate the angle q made
by the rebound velocity with the horizontal and
the velocity v of the balls at the instant they pass
through the opening.
16. A horizontal projectile is fired from a tower B. The
shooter fires his gun from point A at an angle of
30°. Determine the muzzle speed of the bullet if it
hits the projectile at C.
B
A
C
vA
30°
10 m
1.8 m
20 m
17. A particle is projected at an angle of elevation a
and after t second it appears to have an elevation
of b as seen from the point of projection. Find the
initial velocity of projection.
18. A particle is projected with a velocity u at an angle
a to the horizontal, in a vertical plane. At time t, it
is moving in a direction making an angle b with the
horizontal. Prove that gt cos β = u sin( α − β ) .
19. A ball is projected so as to just clear two walls, the
first of height a at a distance b from the point of
projection and the second of height b and at a distance a from point of projection. Show that the
a2 + ab + b2
.
range on the horizontal plane is R =
a+b
Also show that the angle of projection is given by
⎛ a2 + ab + b2 ⎞
tan−1 ⎜
⎟⎠ .
⎝
ab
20. The angular elevation of an enemy’s position on a
hill h m high is b. Show that, in order to shell it, the
minimum initial velocity of the projectile must be
gh ( 1+ cosec β ) .
21. A gun is fired from a moving platform and ranges
of the shot are observed to be R1 and R2 when the
platform is moving forward and backward respectively with velocity v. Find the elevation of the gun
a in terms of the given quantities.
500 mm
θ
400 mm
05_Kinematics 2_Part 1.indd 36
11/28/2019 7:18:32 PM
Chapter 5: Kinematics II
Motion of a Projectile Up an
Inclined Plane
⇒
Consider a projectile to be launched with initial
velocity u making an angle α with horizontal on an
inclined plane which makes angle β(< α ) with the
horizontal. Then u and g can be resolved into two
components each (one along the incline and the other
perpendicular to the incline)
uy
x
A
y
u
B
ax
α –β
ux β
O u cos α
gβ
α
ay
Horizontal
N
ux = u cos (α – β )
uy = u sin (α – β )
ax = –g sin β
ay = –g cos β
OB = Range = R
(a) ux = u cos(α − β ) along the incline along +x axis.
(b) uy = u sin(α − β ) along the incline along +y axis.
(c)
ax = g sin β along −x axis, acting as retardation
to motion along the incline i.e., ax = − g sin β .
(d)
ay = g cos β along −y axis, acting as retardation to motion perpendicular to incline i.e.,
ay = − g cos β .
Range = R =
5.37
2u2 sin ( α − β ) cos α
g cos 2 β
We can also find R by the following method.
1 2
ax t and at t = T , x = R
2
1
⇒ R = ⎡⎣ u cos ( α − β ) ⎤⎦ T + ( − g sin β ) T 2
2
2u sin ( α − β )
Substituting T =
, we get
g cos β
Since, x = ux t +
R=
2u2 sin ( α − β ) cos α
g cos 2 β
Condition for Range to be Maximum
(Up the Incline)
Since, R =
⇒
R=
u2
g cos 2 β
u2
g cos 2 β
⎡⎣ 2 sin ( α − β ) cos α ⎤⎦
⎡⎣ sin ( α − β + α ) + sin ( α − β − α ) ⎤⎦
{∵ 2 sin A cos B = sin ( A + B ) + sin ( A − B ) }
u2
Time of Flight (T ) (Up the Incline)
⇒
Time taken by the particle to go from O to A to B .
When the particle reaches B in a time t = T , then
y = 0.
For R to be MAXIMUM, sin ( 2α − β ) is MAXIMUM
i.e., 1
Since, y = uy t +
1 2
ay t
2
⇒
1
0 = { u sin ( α − β ) } T + ( − g cos β ) T 2
2
⇒
T=
2u sin ( α − β )
g cos β
⇒
Range of Projectile (R ) (Up the Incline)
Since the particle goes from O to N with velocity
component u cos α , uniformly with no acceleration.
So,
ON = ( u cos α ) T
⇒ ON =
2u sin ( α − β ) cos α
g cos β
2
ON
But OB =
cos β
05_Kinematics 2_Part 1.indd 37
g cos 2 β
⎡⎣ sin ( 2α − β ) − sin β ⎤⎦
sin ( 2α − β ) = 1 π
2
π β
⇒ α= +
4 2
Maximum Range is given by
⇒
{∵ T ≠ 0 }
R=
2α − β =
Rmax =
⇒
⇒
⇒
Rmax =
Rmax =
u2
g cos 2 β
( 1 − sin β )
u2
g ( 1 − sin 2 β )
( 1 − sin β )
2
u
g ( 1 + sin β )
Rmax horizontal ⎧
u2 ⎫
Rmax. up =
⎨∵ Rmax horizontal =
⎬
1 + sin β
g ⎭
the incline
⎩
11/28/2019 7:18:43 PM
5.38 JEE Advanced Physics: Mechanics – I
Motion of a Projectile Down an
Inclined Plane
Consider a projectile to be launched with initial
velocity u making a angle α with horizontal down
an inclined plane which makes angle β with the
horizontal. Then u and g can be resolved into two
components each (one along the incline and the other
perpendicular to the incline)
y
uy
u
A
α
ux β
ax
B
gβ
β
ay
Horizontal
N
But OB =
⇒
ux = u cos (α – β )
ux = u sin (α – β )
ax = g sin β
ay = –g cos β
OB = Range = R
(a) ux = u cos ( α + β ) along the incline along +x
axis.
(b) uy = u sin ( α + β ) along the incline along +y
axis.
(c) ax = g sin β along −x axis, acting as acceleration to the motion along the incline.
(d) ay = g cos β along −y axis, acting as retardation to the motion perpendicular to incline i.e.,
ay = − g cos β .
Range = R =
2u2 sin ( α + β ) cos α
g cos 2 β
1 2
ax t and
2
Since, x = ux t +
⇒
R=
⇒
⇒
T=
2u sin ( α + β )
{as T ≠ 0 }
g cos β
Range of Projectile (R) (Down the Incline)
Since the particle goes from O to N with velocity
component u cos α , uniformly with no acceleration.
So,
2u sin ( α + β )
, we get
g cos β
g cos 2 β
Condition for Range to be Maximum
(Down the Incline)
Since, R =
⇒
⇒
1
0 = { u sin ( α + β ) } T + ( − g cos β ) T 2
2
1
( g sin β ) T 2
2
2u2 sin ( α + β ) cos α
Time taken by the particle to go from O to A to B .
When the particle reaches B in a time t = T , then
y = 0.
1
Since, y = uy t + ay t 2
2
R = ⎡⎣ u cos ( α + β ) ⎤⎦ T +
Substituting T =
05_Kinematics 2_Part 1.indd 38
ON
cos β
We can also find R by the following method.
Time of Flight (T ) (Down the Incline)
ON = ( u cos α ) T
2u2 sin ( α + β ) cos α
g cos β
at t = T , x = R
x
O
⇒ ON =
R=
u2
g cos 2 β
u2
g cos 2 β
⎡⎣ 2 sin ( α + β ) cos α ⎤⎦
[sin(α + β + α ) + sin(α + β − α )]
{∵ 2 sin A cos B = sin ( A + B ) + sin ( A − B ) }
R=
u2
g cos 2 β
⎡⎣ sin ( 2α + β ) + sin β ⎤⎦
For R to be MAXIMUM, sin ( 2α + β ) is MAXIMUM
i.e. 1
⇒
sin ( 2α + β ) = 1 π
2
π β
⇒ α= −
4 2
Maximum range is given by
⇒
2α + β =
Rmax =
⇒
Rmax =
u2
g cos 2 β
u2
( 1 + sin β )
g ( 1 − sin 2 β )
( 1 + sin β )
11/28/2019 7:18:52 PM
Chapter 5: Kinematics II
⇒
Rmax =
u2
g ( 1 − sin β )
Since Rmax horizontal =
⇒
Rmax. down =
the incline
u2
g
⇒
2 tan α − 2 tan β = cot β + tan α ⇒
tan α = cot β + 2 tan β …(3)
Condition for the Particle Launched Up
the Incline to Strike it Horizontally
Rmax horizontal
1 − sin β
When the particle strikes the plane horizontally.
Again in this case, time of flight is given by
Condition for the Particle Launched Up
the Incline to Strike it Normally
Let T be the time of flight from O to A , then
T =
5.39
2u sin ( α − β )
…(1)
g cos β
Now we shall consider the motion of the particle
along OA. Initial velocity along OA is
t=
2u sin ( α − β )
g cos β
Since the particle strikes the plane horizontally, so
vertical component of velocity at that instant is zero.
Hence we get
0 = u sin α − gt
⇒
ux = u cos ( α − β )
⎡ 2u sin ( α − β ) ⎤
u sin α = g ⎢
⎥
g cos β
⎣
⎦
A
A
u sin α
u
u
α
α
β
O
Since the particle strikes the plane A at right angles,
so the final velocity along OA is vx = 0.
Acceleration due to gravity along OA is ax = − g sin β
Since x = u2 t +
β
O
1 2
a2 t
2
⇒
sin α cos β = 2 [ sin α cos β − cos α sin β ] ⇒
2 cos α sin β = sin α cos β
⇒
2 tan β = tan α
Problem Solving Technique(s)
⇒
0 = u cos ( α − β ) − ( g sin β ) t ⇒
t=
{∵ vx = ux + ax t }
u cos ( α − β )
…(2)
g sin β
From equations (1) and (2), we get
Following two points are important regarding the
projectile motion in inclined planes:
(a) Time taken by the projectile to move from O to A
is half the time of flight. Here A is a point where
velocity of projectile is parallel to x-direction.
y
T=t
⇒
x
2u sin ( α − β ) u cos ( α − β )
=
g cos β
g sin β
⇒
2 tan ( α − β ) = cot β ⇒
⎛ tan α − tan β ⎞
= cot β
2⎜
⎝ 1 + tan α tan β ⎟⎠
05_Kinematics 2_Part 1.indd 39
A
B
u
O
β α
11/28/2019 7:19:00 PM
5.40 JEE Advanced Physics: Mechanics – I
T u sin( α − β )
i.e., tOA = =
2
g cos β
This can be proved as under,
At A, v y = 0 = uy + a y tOA
Motion Along the Inclined Plane
1
x = v cos(90 − α )t + ( g sin α )t 2 …(2)
2
Motion Perpendicular to the Inclined Plane
⎡ u sin( α − β ) ⎤ u sin( α − β )
⇒ tOA = −
= −⎢
⎥=
ay
g cos β
⎣ − g cos β ⎦
⇒
1
0 = vo sin(90 − α )t + ( − g cos α )t 2
2
(b) At point B, where the projectile strikes the plane
the y-component of its velocity ( v y ) is just equal
and opposite to the component with which it was
projected from point O. i.e., v y = −uy at B
⇒
t=
uy
2vo 2 2 gh
=
g
g
y
v0
α
This can be proved as:
α
h
(90° – α )
v y = uy + a y T
⎡ 2u sin( α − β ) ⎤
v y = u sin( α − β ) + ( − g cos β ) ⎢
⎥
g cos β
⎣
⎦
v = −u sin( α − β ) = −uy
y
in
gs
α
α
α g cos α
g
Put value of t in (2) we get
Illustration 31
A ball starts falling with zero initial velocity on a
smooth inclined plane which forms an angle α with
the horizontal. Having fallen through a height h, the
ball rebounds elastically off the inclined plane. At
what distance from the impact point will the ball
rebound for the second time?
⎛ 4v 2 ⎞
⎛ 2v ⎞ 1
x = vo sin α ⎜ o ⎟ + g sin α ⎜ 2o ⎟
⎝ g ⎠ 2
⎝ g ⎠
⇒
⇒
x=
x=
2v 2o sin α
g
(a) g sin α , acceleration along the inclined plane,
and
(b) g cos α , acceleration perpendicular to the
inclined plane.
After falling freely through a height h, velocity at the
point of striking is
v = 2 gh …(1)
0
Further since the collision is perfectly elastic, hence,
angle of rebound is also a.
05_Kinematics 2_Part 1.indd 40
2vo2 sin α
g
4vo2 sin α
g
Solution
Resolving the component(s) of acceleration w.r.t. the
inclined plane, we have
+
4
( 2 gh ) sin α 2
⇒
x=
⇒
x = 8 h sin α g
{using (1)}
Illustration 32
A large heavy box is sliding without friction down a
smooth inclined plane of inclination θ . From a point
P on bottom of box, a particle is projected inside the
box. The initial speed of particle with respect to box
is u and direction of projection makes and angle α
with the bottom as shown in the figure.
11/28/2019 7:19:05 PM
Chapter 5: Kinematics II
in θ
P u
gs
α
θ
Q
os θ
gc
y
x
u
α
θ
g
5.41
θ
sin
Q
P
θ
(a) Find the distance along the bottom of box
between point of projection P and point Q where
the particle lands. (Assume that the particle does
not hit any other surface of the box. Neglect air
­resistance).
(b) If the horizontal displacement of a particle as
seen by an observer on ground is zero, find the
speed of box with respect to ground at the instant
the particle was projected.
Solution
Lets consider reference frame x-y fixed with box with
origin at one corner O and moving down with box.
The acceleration of this frame down the incline is
g sin θ .
Acceleration of the particle launched inside the
box w.r.t. ground is g (vertically downwards) with
components
(a) g sin θ ; (along the negative x-axis) acting as retardation for particle’s motion along the incline, and
(b) g cos θ ; (along the negative y-axis) acting as
retardation for particle’s motion ­perpendicular
to the incline.
Relative acceleration of the particle in reference
frame along x-axis is
When the particle (strikes) arrives at Q, net displacement perpendicular to the incline is zero.
⇒ 0 = ( u sin α ) t +
⎤
⎡⎛ 1
⎞
⇒ t ⎢ ⎜⎝ 2 g cos θ ⎟⎠ t − ( u sin α ) ⎥ = 0
⎦
⎣
Since t ≠ 0
2u sin α
⇒ t = g cos θ
Since ax = 0 , so horizontal motion is non accelerated motion with uniform velocity, we have
⎛ 2u sin α ⎞
PQ = ( u cos α ) t = ⎜
u cos α
⎝ g cos θ ⎟⎠
u2 sin 2α
⇒ PQ = g cos θ
Method I
If the incline moves down with a velocity v and
moves a distance QP , then it will appear to an
observer on ground that horizontal displacement of
particle is zero
u sin (θ + α )
ax = − g sin θ − ( − g sin θ ) = 0
Relative acceleration of the particle in reference
frame along y-axis is
v cos θ
θ
⇒ ay = − g cos θ
05_Kinematics 2_Part 1.indd 41
u
α
θ
u cos (θ + α )
ay = − g cos θ = 0
Also the x component of particle’s velocity in
box ux = u cos α and the y component of the particle’s velocity in box is uy = u sin α .
1
( − g cos θ ) t 2
2
v sin θ
For Box
PQ = vt +
1
( ax ) t 2
2
11/28/2019 7:19:10 PM
5.42 JEE Advanced Physics: Mechanics – I
⎛ 2u sin α ⎞
2u sin α 1
+ ( g sin θ ) ⎜
g cos θ 2
⎝ g cos θ ⎟⎠
2
⇒
PQ = v
⇒
2u sin α ⎡
u sin θ sin α ⎤ 2u2 sin α cos α
v+
⎥⎦ =
⎢
g cos θ ⎣
cos θ
g cos θ
⇒
⇒
{using part (a)}
sin θ sin α ⎤
⎡
v = u ⎢ cos α −
cos θ ⎥⎦ ⎣
cos ( θ + α )
v=u
cos θ
Method II
For no horizontal displacement,
⇒
u cos ( θ + α ) − v cos θ = 0 ⇒
v=
u cos ( θ + α )
cos θ
Illustration 33
Two bodies are projected from the same point with
equal speeds in such directions that they both strike
the same point on a plane whose inclination is β . If
α be the angle of projection of the first body with the
horizontal show that the ratio of their times of flight
sin ( α − β )
is
.
cos α
Solution
For the first body, we have
u2
R=
{ sin ( 2α − β ) − sin β }
g cos 2 β
Let α ′ be the angle of projection of the second body.
Since, range of both the ­bodies is same. Therefore,
sin ( 2α − β ) = sin ( 2α ′ − β )
⇒
2α ′ − β = π − ( 2α − β ) ⇒ α′ =
Now, T =
⇒
⇒
2u sin ( α − β )
2u sin ( α ′ − β )
and T ′ =
g cos β
g cos β
sin ( α − β )
T
=
=
T ′ sin ( α ′ − β )
sin ( α − β )
⎛π
⎞
sin ⎜ − ( α − β ) − β ⎟
⎝2
⎠
sin ( α − β ) sin ( α − β )
T
=
=
cos α
T′
⎛π
⎞
sin ⎜ − α ⎟
⎝2
⎠
Illustration 34
A particle projected with velocity u strikes at right
angles a plane through the point of projection
inclined at an angle β to the horizon. Show that the
height of the point struck above the horizontal plane
2u2 ⎛ sin 2 β ⎞
through the point of projection is
g ⎜⎝ 1 + 3 sin 2 β ⎟⎠
and that the time of flight up to that instant is,
2u
t=
.
g 1 + 3 sin 2 β
Solution
Time of flight T =
2u sin ( α − β )
…(1)
g cos β
Also, we have, at B , at time t
vx = 0
⇒
ux + ax t = 0 ⇒
u cos ( α − β ) − g sin βt = 0 ⇒
t=
u cos ( α − β )
…(2)
g sin β
These two times being the same, can be equated. So,
we get
T=t
u
u
α
05_Kinematics 2_Part 1.indd 42
π
− (α − β )
2
β
⇒
2u sin ( α − β ) u cos ( α − β )
=
g cos β
g sin β
⇒
2 tan ( α − β ) = cot β ⇒
⎛ tan α − tan β ⎞
= cot β
2⎜
⎝ 1 + tan α tan β ⎟⎠
11/28/2019 7:19:18 PM
5.43
Chapter 5: Kinematics II
y
A
x
cos α =
⇒
sin ( α − β ) = sin α cos β − cos α sin β =
B
u
α
sin β cos β
⇒
1 + 3 sin 2 β
h
β
cos β
1 + 3 sin 2 β
Substituting these values in equation (4), we get
O
⇒
2 tan α − 2 tan β = cot β + tan α
⇒
tan α = 2 tan β + cot β
R=
…(3)
2u2 sin ( α − β ) cos α
g cos 2 β
2u2
cos β
⎛
⎞ ⎛ sin β cos β ⎞
⎜
2
g cos β ⎝ 1 + 3 sin β ⎟⎠ ⎜⎝ 1 + 3 sin 2 β ⎟⎠
⇒
R=
and h = R sin β
⇒
R=
So, let us find R in terms of u and β only by eliminating α , after taking help from equation (3).
So, h = R sin β =
Further OB = Range = R =
2u sin ( α − β ) cos α
2
g cos 2 β
…(4)
1
Since tan α = 2 tan β + cot β = 2 tan β +
tan β
2
2
⇒
2 tan β + 1 1 + sin β
tan α =
=
tan β
sin β cos β
⇒
cos α =
sin β cos β
( 1 + sin 2 β ) + sin 2 β cos2 β
2
2
2u2 sin β
g ( 1 + 3 sin 2 β )
2u2 sin 2 β
g ( 1 + 3 sin 2 β )
Substituting value of sin ( α − β ) =
cos β
1 + 3 sin 2 β
in
(1), we get
T=
2u cos β
2
g cos β 1 + 3 sin β
=
2u
g 1 + 3 sin 2 β
Test Your Concepts-IV
Based on Projectile on an Inclined Plane
1. A heavy particle is projected from a point at the
foot of a fixed plane, inclined at an angle 45° to the
horizontal, in the vertical plane containing the line
of greatest slope through the point. If ϕ ( > 45° ) is
the inclination to the horizontal of the initial direction of projection, for what value of tanf will the
particle strike the plane
(a) horizontally
(b) at right angle
2. Two parallel straight lines are inclined to the horizontal at an angle a. A particle is projected from a
point mid way between them so as to graze one of
the lines and strike the other normally. Show that if
05_Kinematics 2_Part 1.indd 43
(Solutions on page H.150)
q is the angle between the direction of projection
and either of lines, then
tanθ = ( 2 − 1) cot α
3. A projectile is fired with a velocity u at right angles
to the slope, which is inclined at an angle q with the
horizontal. Derive an expression for the distance R
to the point of impact.
4. Determine the horizontal velocity u with which a
stone must be projected horizontally from a point
P, so that it may hit the inclined plane perpendicularly. The inclination of the plane with the horizontal is q and point P is at a height h above the foot of
the incline, as shown in the figure.
11/28/2019 7:19:24 PM
5.44 JEE Advanced Physics: Mechanics – I
P
u
h
horizontal. At the same instant another particle B
is projected with initial velocity u making an angle
b with the horizontal. Both the particles meet again
on the inclined plane. Find the relation between a
and b.
θ
5. A perfectly elastic ball is thrown from the foot of a
plane whose inclination to horizontal is b. If after
striking the plane at a distance R from the point of
projection it rebounds and retraces its former path,
find the velocity of projection.
6. Particle A is released from a point P on a smooth
inclined plane inclined at an angle a with the
05_Kinematics 2_Part 1.indd 44
u
P
B
β
A
α
O
11/28/2019 7:19:24 PM
Chapter 5: Kinematics II
5.45
Solved Problems
The range of the projectile is given by
Problem 1
A particle is moving along a vertical circle of radius
r = 20 m with a constant speed v = 31.4 ms −1 as
shown in figure. Straight line ABC is horizontal and
passes through the centre of the circle. A shell is fired
from point A at the instant when the particle is at C.
If distance AB is 20 3 m and the shell collide with
the particle at B , then prove
tan θ =
3
where n is an integer. Further, show that smallest
value of θ is 30°.
v
u
r = 20 m
θ
20 3 m
B
O
C
Solution
At the time of firing of the shell, the particle was at C
and the shell collides with it at B , therefore the number of revolutions completed by the particle is odd
( 2n − 1 )
multiple of half, i.e.,
, where n is an integer.
2
Let t be the time period of revolution of the particle,
then
t=
From (1) and (2), we get
tan θ =
( 2n − 1 )2
3
⇒ θmin = 30° =
Problem 2
An object A is kept fixed at the point x = 3 m and
y = 1.25 m on a plank P raised above the ground.
At time t = 0, the plank starts moving along the
+x direction with an acceleration 1.5 ms −2 . At the
same instant a stone is projected from the origin
with a velocity u as shown. A stationary person on
the ground observes the stone hitting the object during its downward motion at an angle of 45° to the
horizontal. All the motions are in the x-y plane. Find
u and the time after which the stone hits the object.
Take g = 10 ms −2 .
y
T=
( 2n − 1 )
2
× 4 = 2 ( 2n − 1 ) s
T=
2u sin θ
g
P
u
O
05_Kinematics 2_Part 2.indd 45
3m
x
Solution
For stone to hit the object, A, at time t
y
A
P
1.25 m
45°
u
So, T = t
2u sin θ
= 2 ( 2n − 1 ) …(1)
g
A
1.25 m
For a projectile, the time of flight is given by
⇒
π
radian
3
2π r 2 × 3.14 × 20
=
=4s
v
31.4
If T be the time of flight of the shell, then this time T
⎛ 2n − 1 ⎞
revolutions of the particle
equals the time of ⎜
⎝ 2 ⎟⎠
⇒
2u2 sin θ cos θ
= 20 3 …(2)
g
For smallest θ , n = 1
( 2n − 1 )2
A
R=
O
θ
3m
x
11/28/2019 7:15:49 PM
5.46 JEE Advanced Physics: Mechanics – I
x = (u cos θ )t = xo +
y
1 2
at
2
1
⇒ (u cos θ )t = 3 + (1.5)t 2 …(1)
2
⇒
1
y = (u sin θ )t − gt 2
2
⇒ (u sin θ )t −
1 2
gt = 1.25 …(2)
2
Since stone hits A moving along 45° with horizontal
during its downward motion, so
tan( −45°) =
u sin θ − gt
= −1 …(3)
u cos θ
⇒
u sin θ − gt = −u cos θ ⇒
( u sin θ ) t − gt 2 = − ( u cos θ ) t …(4)
Substituting values of ( u sin θ ) t and ( u cos θ ) t from
(2) and (1) respectively and taking g = 10 ms −2 we
get
1.25 + 5t 2 − 10t 2 = −3 − ( 0.75 ) t 2
v2
v1
O
β
α
(d, 0)
x
For 1st particle, B
⇒ r1 = v1 cos α iˆ + sin α ˆj t
)
(
iˆ
For 2nd particle, A
⇒ r2 = ( d − v2 t cos β ) iˆ + v2 t sin β ˆj
Separation between particles is r = r2 − r1
iˆ
2
2
r = r 2 = ⎡⎣ d − ( v2 cos β + v1 cos α ) t ⎤⎦ +
⎡⎣ ( v2 sin β − v1 sin α ) t ⎤⎦
dr
=0
For distance between particles to be minimum,
dt
dr
= 2 ⎡⎣ d − ( v2 cos β + v1 cos α ) t ⎤⎦ ×
⇒ 2r
dt
2
( −v cos β − v1 cos α ) + 2 ( v2 sin β − v1 sin α )2 t
⇒
( 5 − 0.75 t = ( 3 + 1.25 ) ⇒
t2 = 1 ⇒
t=
⇒
t=1s
⇒
⎡
⎤
d ( v2 sin β − v1 sin α )
⎥
rmin = ⎢
⎢ v 2 + v 2 + 2v v cos ( α + β ) ⎥
1
2
1
2
⎣
⎦
) 2
So, from (1), u cos θ = 3.75 and u sin θ = 6.25
⇒ u = ux iˆ + uy ˆj
⇒ u = ( 3.75 ) iˆ + ( 6.25 ) ˆj iˆ
Problem 3
Two particles A and B simultaneously leave 2 points
O and A separated by a distance d , with velocities
v1 and v2 . The direction in which the particle travels forms an angle α and the second particle forms
an angle β with the line OA. At what time, will the
distance between the particles be minimum, also calculate this minimum distance between particles.
Solution
Method I: FRAME FIXED w.r.t. EARTH
Let distance between the given particles be minimum
at time t
05_Kinematics 2_Part 2.indd 46
)
(
d v2 cos β + v1 cos α
2
[v1 + v22 + 2v1v2 cos(α + β )]
Method II: FRAME FIXED w.r.t. PARTICLE
Since both projectiles are moving under the influence
of gravity having a constant value of g = 9.8 ms −2
acting vertically downwards for both. Hence, we
arrive at the following conclusion(s)
B
v1
C
α
θ
β
v2
v21
O
θ
A(d, 0)
(a) Relative vertical acceleration of one projectile
w.r.t. other is zero.
(b) Relative horizontal acceleration of one projectile
w.r.t. other is zero.
11/28/2019 7:15:57 PM
Chapter 5: Kinematics II
(c) As a consequence of (a) and (b), both the projectiles must follow a straight line motion along AC
with uniform relative velocity v21 , where
v21 = v2 − v1 …(1)
Hence, minimum distance between the particles is
OC i.e. perpendicular distance i.e. length of the perpendicular dropped from O on AC to meet at C.
Further, the angle between v1 and v2 is 180° − ( α + β ) .
Therefore,
Problem 4
The radii of the front and rear wheels of a carriage are
a, b and c is the distance between the axle ­centres.
A particle of dust driven from the highest point of
the rear wheel is observed to alight on the highest
point of the front wheel. Show that the velocity of the
(c + b − a)(c − b + a) g
­carriage is
4(b − a)
Solution
2
v21
= v12 + v22 + 2v1v2 cos ( α + β ) …(2)
Let v21 make an angle θ with the horizontal axis (i.e.
x-axis)
⇒
y = 2b – 2a
c
a
b
)
(
)
(
⇒ v = v ( cos α iˆ + sin α ˆj ) …(5)
⇒
2v
v21 = v21 − cos θ iˆ + sin θ ˆj …(3)
iˆ
v2 = v2 − cos β iˆ + sin β ˆj …(4)
1
For horizontal projectile
y=
1
Substituting (3), (4) & (5) in equation (1) and comparing we get
v21 cos θ = v2 cos β + v1 cos α …(6)
v21 sin θ = v2 sin β − v1 sin α …(7)
⇒
v cos β + v1 cos α
cos θ = 2
v21
⇒
cos θ =
⇒
rmin = OC = d sin θ =
(
)
d ( v2 sin β − v1 sin α )
v12 + v22 + 2v1v2 cos ( α + β )
In right triangle OAC,
rmin = OC = d sin θ =
t=
d ( v2 cos β + v1 cos α )
d cos θ
= 2
v21
[v1 + v22 + 2v1v2 cos(α + β )]
2v 2
2b − 2 a =
⇒
v=
g ⎡⎣ c 2 − (b − a)2 ⎤⎦
2v 2
g(c + b − a)(c − b + a)
4(b − a)
Problem 5
Solution
Lets firstly calculate initial velocity of launch of projectile. Considering origin at O we have
d ( v2 sin β − v1 sin α )
v12 + v22 + 2v1v2 cos ( α + β )
gx 2
A batsman hits a pitched cricket ball at a height of
1.2 m above the ground so that its angle of projection
is 45° and its horizontal range is 110 m. The ball is
lifted towards the left field line where a fence of 7.5 m
is located 100 m from the position of the batsman.
Will the ball clear the fence?
v2 cos β + v1 cos α
2
v1 + v22 + 2v1v2 cos α + β
v
x
iˆ
iˆ
.
Velocity of projection of the dust particle from the top
of the wheel is 2v horizontally.
2
v21
= v12 + v22 − 2v1v2 cos ⎡⎣ 180° − ( α + β ) ⎤⎦
⇒
5.47
−1.2 = 110 tan ( 45° ) −
⇒
g(110)2
2u2 cos 2 ( 45° )
u = 32.66 ms −1 Lets calculate the vertical distance of ball from x-axis
at the position of fence. i.e. we are to find y for
x = 100 m , θ = 45° , u = 32.66 ms −1
05_Kinematics 2_Part 2.indd 47
11/28/2019 7:16:02 PM
5.48 JEE Advanced Physics: Mechanics – I
y
With t as a parameter, these two equations define in
rectangular coordinates the path of point P which
is called a CYCLOID. In other words the two equations give the motion law for the point P on the rim
of the wheel in the case of pure rolling with uniform
motion.
P
u
N
1.2 m
O
y
7.5 m
θ = 45°
A
x
M
100 m
y
Ground
110 m
y = 100 tan ( 45° ) −
C
y P
9.8 × (100)2
O
2 × (32.66)2 cos 2 ( 45° )
y = 100 − 91.87 = 8.13 m
Total height of ball from the ground is
⎛ v t⎞
vy = vo sin ⎜ o ⎟ …(2)
⎝ R ⎠
Problem 6
R
C
O P
v0
x
⇒
⎡
⎛ v t⎞⎤
v = vx2 + vy2 = v0 2 ⎢ 1 − cos ⎜ 0 ⎟ ⎥
⎝ R ⎠⎦
⎣
⇒
⎛ v t⎞
v = 2v0 sin ⎜ 0 ⎟ …(3)
⎝ 2R ⎠
From this expression, we see that the maximum
πR
speed of point P is 2v0 when t =
, that is, when
vo
the point P is at the top of its path.
Differentiating equation (2) again with respect to
time, we obtain the acceleration-time equations
⇒
ax =
vo2
⎛ v t⎞
sin ⎜ o ⎟
⎝ R ⎠
R
ay =
vo2
⎛ v t⎞
cos ⎜ o ⎟ …(4)
⎝ R ⎠
R
Solution
After the time interval t , the center C of the wheel
will have travelled a distance v0 t as shown, and since
it rolls without slipping, the arc DP will also have
vt
the length v0 t . Thus the angle DCP will be θ = o ,
R
Then from the geometry of the figure, we can express
the coordinates x and y of the point P as follows:
⎛ v t⎞
x = vo t − R sin ⎜ o ⎟
⎝ R ⎠
⎛ v t⎞
y = R − R cos ⎜ o ⎟ …(1)
⎝ R ⎠
05_Kinematics 2_Part 2.indd 48
x
D
⎡
⎛ v t⎞⎤
vx = vo ⎢ 1 − cos ⎜ o ⎟ ⎥
⎝ R ⎠⎦
⎣
Since 9.33 m > 7.5 m , therefore, the ball will clear the
fence.
y
x
v0
Differentiating equation (1) with respect to time gives
the velocity-time equations as follows:
( 1.2 + 8.13 ) = 9.33 m
A wheel of radius R rolls without slipping along the
x-axis with constant speed v0 (as shown in the fi
­ gure).
Investigate the motion of a point P on the rim of the
wheel which starts from the origin O and find the
total distance covered by the point between two successive moments at which it touches the surface.
θ R
Therefore, a = ax2 + ay2 =
vo2
…(5)
R
Thus the point P has acceleration of constant magnitude always directed towards the centre C of the
rolling wheel.
The total distance traversed by the point P
between two successive moments at which it touches
the surface.
11/28/2019 7:16:08 PM
Chapter 5: Kinematics II
⇒
⇒
s=
∫
2π R
vo
vdt =
∫
0
⎡
⎛ vo t ⎞ ⎤
⎢ 2vo sin ⎜⎝ 2R ⎟⎠ ⎥ dt
⎦
⎣
s = 8R ⇒
Problem 7
A canon fires successively two shells with velocity u,
the first at an angle θ1 and second at an angle θ 2 with
horizontal. Neglecting air drag. Find the time interval
between firings leading to collision of shells.
Solution
METHOD I
Let the time of flight before collision be t, i.e., let
first shell reach P ( x , y ) in time t .
Let second shell be fired with a delay say Δt.
So it falls short of time by Δt and has net time ( t − Δt )
to be at P.
For First shell
⇒
⇒
gx ⎡ cos 2 θ 2 − cos 2 θ1 ⎤
sin θ1 sin θ 2
−
= 2⎢
⎥
cos θ1 cos θ 2 2u ⎣ cos 2 θ 2 × cos 2 θ1 ⎦
sin ( θ1 − θ 2 )
cos θ1 + cos θ 2
Δt =
x = ( u cos θ 2 ) ( t − Δt ) …(2)
⇒
⇒
t − Δt cos θ1
=
cos θ 2
t
Δt cos θ 2 − cos θ1
=
…(3)
cos θ 2
t
From (1), t =
⇒
⇒
x
u cos θ1
Δt ( u cos θ1 )
x
cos θ 2 − cos θ1
=
cos θ 2
gx 2
2u2 cos 2 θ1
gx 2
⇒
y = x tan θ 2 −
⇒
tan θ1 − tan θ 2 =
05_Kinematics 2_Part 2.indd 49
2
2
2u cos θ 2
METHOD II
Since no friction is there, so let us assume both shells
are fired at same instant. Let they pass through common point P ( x , y ) at times t1 and t2 from the start,
i.e., they miss each other by time ( t1 − t2 ) and hence,
if this is the time lapse between firing of shells then
they will collide at P ( x , y )
y
P(x, y)
θ1
θ2
x
O
For First Shell
x = ( u cos θ1 ) t1 …(1)
1
⎛
⎞
y = ⎜ u sin θ1t1 − gt12 ⎟ …(2)
⎝
⎠
2
Also
For Second Shell
x = ( u cos θ 2 ) t2 …(3)
uΔt cos θ 2 − cos θ1
=
…(4)
x
cos θ1 × cos θ 2
Also, y = x tan θ1 −
gx ⎛ uΔt ⎞
⎜
⎟
2u 2 ⎝ x ⎠
2u ⎡ sin ( θ1 − θ 2 ) ⎤
⎢
⎥
g ⎣ cos θ1 + cos θ 2 ⎦
x = ( u cos θ1 ) t …(1)
For Second shell
=
5.49
…(5)
…(6)
gx ⎡ 1
1 ⎤
−
⎥
2 ⎢
2
2u ⎣ cos θ1 cos 2 θ 2 ⎦
{using (5) & (6)}
u ( sin θ 2 t2 − sin θ1t1 ) =
1
g ( t1 + t2 ) ( t1 − t2 ) …(4)
2
Using relations (1) and (3), we get
⎛
cos θ 2 ⎞
u ⎜ sin θ 2 t2 − sin θ1
t2 =
cos θ1 ⎟⎠
⎝
1 ⎛ cos θ1 + cos θ 2 ⎞
g
⎟⎠ t2 Δt …(5)
cos θ1
2 ⎜⎝
Also using relations (3) and (4), we get
u sin θ 2 t2 −
1 2
1
gt2 = u sin θ1t1 − gt12
2
2
11/28/2019 7:16:15 PM
5.50 JEE Advanced Physics: Mechanics – I
⇒
u ( sin θ 2 t2 − sin θ1t1 ) =
Using (5), we get
1
g ( t1 + t2 ) ( t1 − t2 )
2
⎛
cos θ 2 ⎞
u ⎜ sin θ 2 t2 − sin θ1
t2 =
cos θ1 ⎟⎠
⎝
1 ⎛ cos θ1 + cos θ 2 ⎞
g
⎟⎠ t2 Δt
cos θ1
2 ⎜⎝
⇒
2u ⎡ sin ( θ1 − θ 2 ) ⎤
Δt =
⎥
⎢
g ⎣ cos θ1 + cos θ 2 ⎦
(b) Time of flight,
A train is moving with a constant speed of 10 ms −1
16
m. The plane of the circle lies
in a circle of radius
π
in horizontal x -y plane. At time t = 0 train is at point
P and moving in counter-clockwise direction. At this
instant a stone is thrown from the train with speed
10 ms −1 relative to train towards negative x-axis at
an angle of 37° with vertical z-axis. Find
lands on the ground are ( −4.5 m, 16 m, 0 )
At highest point, we have
t=
T
= 0.8 s
2
⇒
x=
16
− ( 6 )( 0.8 ) = 0.3 m
π
⇒
y = ( 10 )( 0.8 ) = 8 m
and z =
x
P
(a) the velocity of particle relative to train at the
highest point of its trajectory.
(b) the co-ordinates of points on the ground where
it finally falls and that of the highest point of its
trajectory.
Therefore, coordinates at highest point are
( 0.3 m, 8 m, 3.2 m ) .
A particle is projected from an inclined plane OP1
from A with velocity v1 = 8 ms −1 at an angle 60°
with horizontal. An another particle is projected at
the same instant from B with velocity 16 ms −1 and
perpendicular to the plane OP2 as shown in figure.
After time 10 3 s separation between them was
minimum and found to be 70 m. Find the
v1
P1
A
v2
60°
90°
45°
Take g = 10 ms , sin ( 37° ) = 0.6
−2
P2
B
30°
O
(a) distance AB.
(b) height of A and B from O
( )
At t = 0, vT = 10 ˆj ms −1
vST = 10 cos 37°kˆ − 10 sin 37° ˆj = ( 8 kˆ − 6iˆ ) ms −1
Since, vST = vS − vT
⇒ vS = vST + vT = −6iˆ + 10 ˆj + 8 kˆ ms −1 iˆ
(
Solution
( v21 )x = ( v1 + v2 ) cos 60° = 12 ms −1
)
(a) At highest point, vertical component ( k̂ ) of vS
will become zero. Hence, velocity of particle at
highest point will become −6iˆ + 10 ˆj ms −1
(
05_Kinematics 2_Part 2.indd 50
2
vZ2 ( 8 )
=
= 3.2 m
2g
20
Problem 9
y
Solution
2vZ 2 × 8
=
= 1.6 s
g
10
⇒ y = ( 10 )( 1.6 ) = 16 m and z = 0
Therefore coordinates of particle where it finally
Problem 8
T=
)
( v21 )y = ( v1 + v2 ) sin 60° = 4 3 ms −1
⇒
2
v21 = ( 12 ) + ( 4 3 ) = 13.9 ms −1 2
11/28/2019 7:16:22 PM
Chapter 5: Kinematics II
Solution
v21
The coordinates of the points A, B, C and D are
­calculated here for convenience.
C
y
y
A
θ
45°
105°
O
α
30°
x
B
B
Applying sine law in ΔABO, we get
a
2
x
x
a
⎛
B⎜ x + ,
⎝
2
⎞
3a ⎟
⎠
3a
⎛
C⎜ x + ,
⎝
2
⎞
3a ⎟
⎠
⎛
D ⎜ x + 2 a,
⎝
3a ⎞
⎟
2 ⎠
a
a
+ a + + x = 2 a + 2x
2
2
Also we know that
x⎞
⎛
y = x tan θ ⎜ 1 − ⎟
⎝
R⎠
BO
sin ( 180° − ( α + 30° + 105° ) )
Since A lies on Trajectory, so
AO = 181.2 m and BO = 134.6 m
So, hA = AO sin 45° = 128 m and
3a
x ⎞
⎛
= x tan θ ⎜ 1 −
⎟ …(1)
⎝
2
2 a + 2x ⎠
Also, B lies on Trajectory, so
hB = BO sin 30° = 67.3 m
a⎞ ⎞
⎛
⎛
⎜⎝ x + ⎟⎠ ⎟
⎜
a⎞
⎛
2
3 a = ⎜ x + ⎟ tan θ ⎜ 1 −
⎟ …(2)
⎝
⎝
2⎠
2 a + 2x ⎠
Problem 10
A particle is launched such that it grazes the four vertices of a regular hexagon of side a as shown. Find
the range of the particle.
Dividing (1) and (2), we get
3
a
2 =
3a
y
a
05_Kinematics 2_Part 2.indd 51
a
Range, R = x +
AB
AO
=
=
sin ( 105° ) sin ( α + 30° )
a
a
a
a
a
a
⎛
3 ⎞
A ⎜ x,
x⎟
⎝
2 ⎠
−1 ⎛ 70 ⎞
⇒ ∠α = 30° − sin ⎜⎝ 250 ⎟⎠ = 13.7°
O
D
θ 60°
O x a
2
⎛ AC ⎞
(b) ∠α = ∠θ − ∠ABC = 30° − sin −1 ⎜
⎝ AB ⎟⎠
a
a
a sin 60°
2
2
AB = ( 240 ) + ( 70 ) = 250 m
Solving this we get,
C
A
(a) Hence using Pythagora′s Theorem, we have
a
a
30°
⎛4 3⎞
and θ = tan −1 ⎜
= 30° , BC = ( v21 ) t = 240 m and
⎝ 12 ⎟⎠
AC = 70 m {given}
5.51
x
⇒
x ⎞
⎛
x tan θ ⎜ 1 −
⎟
⎝
2 a + 2x ⎠
a⎞ ⎞
⎛
⎛
⎜⎝ x + ⎟⎠ ⎟
⎜
a⎞
⎛
2
⎜⎝ x + ⎟⎠ tan θ ⎜⎝ 1 −
⎟
2
2 a + 2x ⎠
a⎞ ⎛
a⎞
⎛
⎜⎝ x + ⎟⎠ 2 a + 2x − x −
⎜
2
2⎟
2=
⎜⎝
⎟
x
2 a + 2x − x ⎠ 11/28/2019 7:16:27 PM
5.52 JEE Advanced Physics: Mechanics – I
⇒
⇒
⇒
For a given value of x , maximum y can be determined from
⎛ 3a
⎞
+x
⎟
⎛ 2x + a ⎞ ⎜ 2
2=⎜
⎝ 2x ⎟⎠ ⎜⎝ 2 a + x ⎟⎠
dy
1 ⎛ 2x + a ⎞ ⎛ 3 a + 2x ⎞
2= ⎜
⎟⎜
⎟
4 ⎝ x ⎠ ⎝ 2a + x ⎠
d ( tan θ )
(
)
(
)
(
8 x 2 a + x = 2x + a 3 a + 2x ) ⇒
x−
gx 2
2u2
⇒ 16 ax + 8 x 2 = 4 x 2 + 6 ax + 2 ax + 3 a 2 ⇒
⇒
⇒
⇒
⇒
2
2
4 x + 8 ax − 3 a = 0 2
⇒
2x + 2 a = a 7 ⇒
R = 7a tan θ =
u2
gx
On substituting the expression for tan θ in equation (1), we get
a
x = −a ±
7
2
a
x+a=
7
2
⇒
( 2 tan θ ) = 0
⎧ d ( tan 2 θ )
⎫
d ( tan θ )
= 2 tan θ and
= 1⎬
⎨∵
d ( tan θ )
d ( tan θ )
⎩
⎭
2
8 a ± 64 a + 48 a
8
8 x = −8 a ± 4 a 4 + 3 x=−
=0
⎛ u2 ⎞ gx 2 ⎛
u4 ⎞
y MAX = x ⎜
−
+
1
⎜
⎟
⎝ gx ⎟⎠ 2u2 ⎝
g2x2 ⎠
⇒
y MAX =
u2 gx 2
−
2 g 2u2
The shell can hit an area defined by
Problem 11
y ≤ y MAX
An enemy fighter jet is flying at a constant height
of 250 m with a velocity of 500 ms −1 . The fighter jet
passes over an anti-aircraft gun that can fire at any
⇒
time and in any direction with a speed of 100 ms −1.
Determine the time interval during which the fighter
jet is in danger of being hit by the gun bullets.
On substituting numerical values,
u = 100 ms −1, g = 10 ms −2 , we get
The equation of trajectory of bullets is
⎛ gx 2 ⎞
y = x tan θ − ⎜ 2 ⎟ sec 2 θ
⎝ 2u ⎠
⇒
y = x tan θ −
2u2
u2 gx 2
−
2 g 2u2
y = 250 m,
x2
≤ 250
2000
Solution
gx 2
y≤
( 1 + tan 2 θ ) …(1)
y
⇒
x 2 ≤ 500000 ⇒
−500 2 ≤ x ≤ 500 2 The fighter jet, can travel 1000 2 m while it can be
hit. So the plane is in danger for a period of
t=
Danger zone
x 1000 2
=
= 2 2 s.
u
500
250 m
–500 2
05_Kinematics 2_Part 2.indd 52
+500 2
x
11/28/2019 7:16:34 PM
Chapter 5: Kinematics II
5.53
Practice Exercises
Single Correct Choice Type Questions
This section contains Single Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONLY ONE is correct.
3
(B) tan −1 ⎛⎜ ⎞⎟
(A) 30°
1. Two particles P and Q are moving as shown in the
⎝ 4⎠
figure. At this moment of time the angular speed of P
(C) 45°
(D) 60°
w.r.t. Q is
30°
P
Q
2.5 m
(A) 5 rads −1
(C) 2 rads −1
2.
4.
(
(B) 4 rads −1
(D) 1 rads −1
y
θ
5.
x
v
a
(A) h =
2v 2 sin θ cos θ
a
h=
2v 2 sin θ cos θ
g
(B)
3.
h
(C) h =
a
2v 2
⎛
⎞
sin θ ⎜ cos θ + sin θ ⎟
g
g
⎝
⎠
(D) h =
g
2v 2
⎛
⎞
sin θ ⎜ cos θ + sin θ ⎟
⎝
⎠
a
a
The speed of a projectile when it is at its greatest height
2
times its speed at half the m
­ aximum height. The
is
5
angle of projection is
05_Kinematics 2_Part 2.indd 53
)
θ
A particle is ejected from the tube at A with a velocity
v at an angle θ with the vertical y-axis. A strong horizontal wind gives the particle a constant horizontal
acceleration a in the x-direction. If the particle strikes
the ground at a point directly under its released position and the downward y-acceleration is taken as g
then
A
A boy throws a ball upwards with velocity
v0 = 20 ms −1 . The wind imparts a horizontal acceleration of 4 ms −2 to the left. The angle θ at which the ball
must be thrown so that the ball returns to the boy’s
hand is g = 10 ms −2
6.
7.
v0
Wind
60°
8 ms–1
3 ms–1
(A) tan −1 ( 1.2 )
(B)
(C) cot −1 ( 2 )
(D) cot −1 ( 2.5 )
tan −1 ( 0.2 )
A particle is moving in a circle of radius R in such a
way that at any instant the total acceleration makes an
angle of 45° with radius. Initial speed of particle is v0 .
The time taken to complete the first revolution is
(A)
R
v0
(B)
2R
v0
(C)
R −2π
e
v0
(D)
R(
1 − e −2π )
v0
A large number of bullets are fired in all the directions
with the same speed v . The maximum area on the
ground on which these bullets will spread is
(A)
π v2
g
(B)
π v4
g2
(C)
π 2v 4
g2
(D)
π 2v 2
g2
Starting from rest, a particle rotates in a circle of radius
π
R = 2 m with an angular acceleration α = rads −2 .
4
The magnitude of average ­velocity of the particle over
the time it rotates quarter ­circle is
11/28/2019 7:16:41 PM
5.54 JEE Advanced Physics: Mechanics – I
8.
(A) 1 ms −1
(B) 1.25 ms −1
(A) 23.5 ms −1
(B)
(C) 1.5 ms −1
(D) 2 ms −1
(C) 47 ms −1
(D) 32.5 ms −1
A projectile has a horizontal range R for two different angles. If h1 and h2 are the maximum heights
reached, then
(A) R = h1h2
(C) R = 4 h1h2
9.
R = h1h2
(B)
(D) R = 2 h1h2
A hollow vertical cylinder of radius r and height
h has a smooth internal surface. A small particle is
placed in contact with the inner side of the upper rim,
at point A and given a horizontal speed u , tangential
to the rim. It leaves the lower rim at point B , vertically
below A . If n is an ­integer then
A
u
12. A ball is thrown vertically upward with a speed v
from a point h metre above the ground. The time
taken for the ball to hit the ground is
2 hg ⎞
⎛ v⎞⎛
(A) ⎜ ⎟ ⎜ 1 + 1 + 2 ⎟
v ⎠
⎝ g⎠⎝
(B)
⎛ 2 hg ⎞
⎛ v⎞
(C) ⎜ ⎟ 1 + ⎜ 2‘ ⎟
⎝ v ⎠
⎝ g⎠
⎛ 2 hg ⎞
⎛ v⎞
(D) ⎜ ⎟ 1 − ⎜ 2 ⎟
⎝ v ⎠
⎝ g⎠
⎛ 2 hg ⎞
1+ ⎜ 2 ⎟
⎝ v ⎠
13. The muzzle velocity for a certain rifle is 600 ms −1 .
If the rifle is pointed vertically upward and fired from
an automobile moving horizontally at a speed of
72 kmh −1 , the radius of curvature of the path of the
bullet at maximum altitude is (neglect friction of the
air and take g = 10 ms −2 )
(A) 400 m
(C) 40 m
h
235 ms −1
(B) 200 m
(D) 20 m
14. A particle starts from the origin of co-ordinates at time
t = 0 and moves in the xy plane with a constant acceleration α in the y-direction. Its equation of motion is
y = β x 2 . Its velocity component in the x-direction is
B
r
(A)
u 2h
=n
2π r g
(B)
h
=n
2π r
(A)
α
2β
(B)
(C)
2π r
=n
h
(D)
u
=n
2 gh
(C)
2α
β
(D) not constant
10. A ball is projected so as to pass a wall at a distance a
from the point of projection at an angle of 45° and
falls at a distance b on the other side of the wall. If h
is the height of the wall then
(A) h = a 2
(C) h =
(B)
2 ab
a+b
h=b 2
(D) h =
ab
a+b
11. A boy wants to throw a ball from a point A so as to
just clear the obstruction at B . The minimum horizontal velocity with which the boy should throw the ball is
A
30 m
u
B
20 m
12 m
05_Kinematics 2_Part 2.indd 54
α
2β
15. The speed of a projectile u reduces by 50% on reaching maximum height. The range on the h
­ orizontal
plane is
(A)
2u2
3g
(B)
u2
g
(C)
3u 2
2g
(D)
3u 2
g
16. It is observed that a projectile is at the same height at
3 s and 5 s from the start. The time of flight of the
projectile is equal to
(A) 1 s
(B) 2 s
(C) 4 s
(D) 8 s
17. A jet plane flying at a constant velocity v at a height
h = 8 kilometre is being tracked by a radar R located
at O directly below the line of flight. If the angle θ is
decreasing at the rate of 0.025 rads −1 , the velocity of
the plane when θ = 60° is
11/28/2019 7:16:51 PM
Chapter 5: Kinematics II
22. A particle is projected with a certain velocity at an
angle α above the horizontal from the foot of an
inclined plane of inclination 30° . If the particle strikes
the plane normally then α is equal to
v
r
h
θ
O
R
(A) 1440 kmh −1
(B)
(C) 1920 kmh −1
(D) 480 kmh −1
5.55
960 kmh −1
18. A helicopter ascending at the rate of 12 ms −1 drops a
food packet from a height of 80 m above the ground.
The time the packet takes to reach the ground is
(B) 5.5 s
(A) 2.7 s
(C) 11 s
(D) 16.5 s
19. A ball is thrown from the top of a staircase which just
touches the ceiling and finally hits the bottom of the
steps. The initial speed of the ball is
2m
⎛ 1 ⎞
(A) 30° + tan −1 ⎜
⎝ 2 3 ⎟⎠
(B)
(C) 60°
⎛ 3⎞
(D) 30° + tan −1 ⎜
⎝ 2 ⎟⎠
30° + tan −1 ( 2 3 )
23. A particle moves along a parabolic path y = 9x 2 in
such a way that the x component of velocity remains
1
constant and has a value ms −1 . The acceleration of
3
the particle is
(A)
1
j ms −2
3
(B)
(C)
2
j ms −2
3
(D) 2j ms −2
3 j ms −2
24. A particle moves along a circle of radius R = 2 m so
that its radius vector r relative to a point on its circumference rotates with the constant angular velocity
ω = 2 rads −1 . The linear speed of the particle is
θ /2 R
3m
θ /2
R
θ
4m
(A) 6.3 ms −1
(C) 63 ms
(B)
−1
7.6 ms −1
(D) 76 ms −1
−1
20. A stone is projected with a velocity of 10 ms at 60°
to the horizontal. At any instant, the angles of elevation of the stone from the two extremities of the range
are α and β. Then tan α + tan β equals
1
3
(A)
(B)
(C) 3
(D)
1
3
3
21. A particle is projected from the ground with a speed
of 20 ms −1 making an angle of 60° with the horizontal. The radius of curvature of the path of the particle,
when its velocity makes an angle of 30° with horizontal is g = 10 ms −2
(
(A) 10.6 m
(C) 15.4 m
05_Kinematics 2_Part 2.indd 55
)
(B) 12.8 m
(D) 24.2 m
(A) 4 ms −1
(B)
2 ms −1
(C) 1 ms −1
(D) 8 ms −1
25. For two projectiles launched with same initial velocity,
the maximum heights corresponding to equal ranges
are 4 m and 16 m . The range has a value
(A) 4 m
(B) 8 m
(C) 16 m
(D) 32 m
26. When a particle is moving along a circular path with
uniform speed, it has
(A) radial velocity and radial acceleration
(B) radial velocity and transverse acceleration
(C) transverse velocity and radial acceleration
(D)transverse velocity and transverse acceleration
27. A particle moves in the xy plane with constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y = α x − β x 2 ,
11/28/2019 7:16:58 PM
5.56 JEE Advanced Physics: Mechanics – I
where a and β are positive constants. The velocity of
the particle at the origin of coordinates is
(A)
⎛ α2 + 1⎞
a⎜
⎝ 2β ⎟⎠
(B)
⎛ β + 1⎞
a⎜
⎝ 2α 2 ⎠⎟
(C)
⎛ α2 + 1⎞
a⎜
⎝ 4β ⎟⎠
(D)
⎛ β + 1⎞
a⎜
⎝ 4α 2 ⎟⎠
28. A body of mass m thrown horizontally with velocity v from the top of the tower of height h touches
the ground at a distance of 250 m from the foot of the
tower. A body of mass 2m thrown with a velocity of
v
from the top of the tower of height 4h will touch
2
the ground at a distance of
(A) 250 m
(B) 500 m
(C) 1000 m
(D) 125 m
29. Two particles A and B are connected by a rigid
rod AB . The rod slides along perpendicular rails as
shown here. The velocity of A to the left is 10 ms −1 .
The speed of B when angle θ = 45° is
B
A
(A) 5 ms −1
(C) 7.5 ms −1
(B) 5 2 ms −1
(D) 10 ms −1
30. A stone is projected so as to pass two walls of heights
a and b at distances b and a respectively from the
point of projection. If α is the angle of projection then
respectively. The time interval between their passing
through the other common point of their path (other
than origin) is
(A)
2 ⎛ v1u1 − v2u2 ⎞
g ⎜⎝ u1 + u2 ⎟⎠
(B)
2 ⎛ v12 + v22 ⎞
g ⎜⎝ u1 + u2 ⎟⎠
(C)
2 ⎛ u12 + u22 ⎞
g ⎜⎝ v1 + v2 ⎟⎠
(D)
2 ⎛ v1u2 − v2u1 ⎞
g ⎜⎝ u1 + u2 ⎟⎠
33. A pendulum of length l = 1 m is released from
θ 0 = 60° . The rate of change of speed of the bob at
θ = 30° is g = 10 ms −2
(
60°
(A) 2.5 ms −2
(B)
(C) 5 3 ms −2
(D) 10 ms −2
32. Two particles are projected from a point at the same
instant with velocities whose horizontal components
and vertical components are ( u1 , v1 ) and ( u2 , v2 ) ,
05_Kinematics 2_Part 2.indd 56
5 ms −2
34. A projectile is thrown in a viscous medium offering
resistance equal to one-tenth of acceleration due to
gravity. The time of flight of projectile will
(A) increase by 1%
(B) decrease by 1%
(C) increase by 2%
(D) decrease by 2%
35. A particle is projected under gravity with velocity
2ag from a point at a height h above the level plane.
The maximum range R on the ground is
v=
(A) minimum value of tan α is 3 .
(B) minimum value of tan α is 3.
(C) maximum value of tan α is 3 .
(D) maximum value of tan α is 3.
31. Two stones are projected so as to reach the same distance from the point of projection on a horizontal surface. The maximum height reached by one exceeds
the other by an amount equal to half the sum of the
heights attained by them. Then the angles of projection
for the stones are
(B) 0°, 90°
(A) 45°, 135°
(D) 20°, 70°
(C) 30°, 60°
)
O
2ag
θ
h
P
R
(A)
( a2 + 1 ) h
(B)
(C)
ah
(D) 2 a ( a + h )
a2 h
36. The position vector of particle, moving in x -y plane,
at any time t is r = ⎡⎣ ( 2t ) i + ( 2t 2 ) j ⎤⎦ m . If θ be the
angle which its velocity vector makes with positive
x-axis) then the rate of change of θ at time t = 0.5 s .
11/28/2019 7:17:06 PM
Chapter 5: Kinematics II
(A) 6 rads −1
(C) 2 rads
(B)
−1
4 rads −1
(D) 1 rads
−1
37. Two particles are projected simultaneously in the same
vertical plane, from the same point, but with different
speeds and at different angles with the horizontal. The
path followed by one, as seen by the other, is
(A) a vertical straight line.
(B)a straight line making a constant angle ( ≠ 90° )
with the horizontal.
(C) a parabola.
(D) a hyperbola.
38. A particle P is sliding down a frictionless hemispherical bowl. It passes the point A at t = 0 . At this instant
of time, the horizontal component of its velocity is v .
A bead Q of same mass as P is ejected from A at
t = 0 along the horizontal string AB , with a speed
v . Friction between the bead and the string may be
neglected. Let tP and tQ be the respective times taken
by P and Q to reach the point B , then
Q
A
B
P
C
(B)
Length of arc ACB
tP
=
tQ Length of chord AB
39. For the simple pendulum shown, l = 200 mm , when
dθ
θ = 30° ,
= −9 rad s −1 . The magnitude of the total
dt
acceleration of the pendulum for this position is
θ
(A) 9.81 ms −2
(C) 16.2 ms
B
C
A
−2
(B)
v
x
D
(A)
v
2
(B)
v
2
(C)
v
3
(D)
v
4
42. The distance r from the origin of a particle moving in x -y plane varies with time as, r = 2t and the
angle made by the radius vector with positive x-axis
is θ = 4t . Here, t is in second, r in metres and θ in
radian. The speed of the particle at t = 1 s is
(B) 10 ms −1
−1
(D) 8 ms −1
43. A car enters a curved road in the form of a quarter of a
circle, the path length being 200 metre . Its speed at the
entrance is 18 kmh −1 but when it leaves, it increases to
54 kmh −1 . If the car is travelling with constant acceleration along the curve, the acceleration when the car
leaves the curved road is
54 kmh–1
l
18 kmh–1
5 ms −2
(D) 17 ms
−2
40. A bomber flying with a horizontal velocity of
500 kmh −1 at a vertical height of 5 km above the
ground wants to hit a train moving with a velocity of
100 kmh −1 in the same direction and in the same vertical plane. The angle θ between the line of sight of the
target and the horizontal at the instant the bomb shell
should be released is approximately
05_Kinematics 2_Part 2.indd 57
41. Four rods each of length l have been hinged to form a
rhombus. Vertex A is fixed to rigid support, vertex C
is being moved along the x-axis with a constant velocity v as shown in the ­figure. The rate at which vertex
B is approaching the x-axis at the moment the rhombus is in the form of a square is
(C) 8.25 ms
tP = tQ
(C) tP > tQ
(D)
(B) 57°
(D) 27°
(A) 55°
(C) 72°
(A) 12 ms −1
(A) tP < tQ
5.57
(A) 0.9 ms −2
(B)
0.45 ms −2
(C) 1.9 ms −2
(D) 3.68 ms −2
44. The magnitude of displacement of a particle moving
in a circle of radius a with constant angular speed ω
varies with time t as
(A) 2a cos ( ωt )
(B) 2a sin ( ωt )
⎛ ωt ⎞
(C) 2 a cos ⎜
⎝ 2 ⎟⎠
⎛ ωt ⎞
(D) 2 a sin ⎜
⎝ 2 ⎟⎠
11/28/2019 7:17:17 PM
5.58 JEE Advanced Physics: Mechanics – I
45. Two bodies are thrown simultaneously from the same
point. One thrown straight up and the other at an
angle α with the horizontal. Both bodies have velocity equal to v0 . Neglecting the air drag, the separation
between the particles at time t is
(A) 2v0t 1 − cos α
(B)
v0t 1 − cos α
⎛π α⎞
(C) 2v0t sin ⎜ − ⎟
⎝ 4 2⎠
⎛π α⎞
(D) 2v0t cos ⎜ − ⎟
⎝ 4 2⎠
49. A particle A is projected from the ground with an initial velocity of 10 ms −1 at an angle of 60° with horizontal. At the same instant, another particle B is projected horizontally with velocity 5 ms −1 from height
h above A so that both the particles collide at point
C on the ground. Taking g = 10 ms −2 , then
46. A boat is moving directly away from the gun on the
shore with speed v1 . The gun fires a shell with speed
v2 at an angle α and hits the boat. The ­distance of the
boat from the gun at the moment it is fired is
(A)
2v2 sin α ( v1 cos α − v2 )
g
(B)
2v1 sin α ( v1 cos α − v2 )
g
(C)
2v2 sin α ( v2 cos α − v1 )
g
(D)
h
10 ms–1
60°
A
(A) h = 10 m
(C) h = 15 m
C
(B) h = 30 m
(D) h = 25 m
50. A body is projected with a velocity u at an angle α
with the horizontal. It passes over a wall at a distance
x from the point of projection. The maximum height
of the wall corresponds to
2v2 cos α ( v1 sin α − v2 )
g
47. A projectile is thrown into space so as to have maximum possible range of 400 m. Taking the point of projection as the origin, the co-ordinate of the point where
the velocity of the projectile is minimum is
(A) (400, 100)
(B) (200, 100)
(C) (400, 200)
(D) (200, 200)
48. A number of projectiles each with a fixed muzzle
velocity u are fired at different angles lying between
0° and 90° as shown. Neglecting air resistance and
assuming g to be constant, then the equation of the
envelope E of the parabolic trajectories (as shown in
figure) is
y
E
A
x
(A) y =
u2 gx 2
−
2 g 2u2
(B)
y=
gx 2 u2
+
2u 2 2 g
(C) y =
2u 2 2 g
−
gx 2 u2
(D) y =
2u 2 2 g
+
gx 2 u2
05_Kinematics 2_Part 2.indd 58
5 ms–1
B
(A) tan α =
u
gx
(B)
tan α =
u2
2 gx
(C) tan α =
u2
gx
(D) tan α =
u2
2g
51. In PROBLEM 50, the maximum height of the wall is
(A) h =
u2
2g
(B)
(C) h =
u2 gx 2
+
2 g 2u2
(D) h =
h=
gx 2
2u2
u2 gx 2
−
2 g 2u2
52. Two bodies are projected simultaneously in the same
vertical plane, from the same point with different
speeds and at different angles of projection with the
horizon. The path followed by one projectile as seen
from the other is
(A) a horizontal straight line.
(B) a parabola.
(C) a hyperbola.
(D)a straight line inclined at an acute angle with the
horizontal.
53. Two particles are projected from the ground simulta10
ms −1 at angles
neously with speeds 10 ms −1 and
3
30° and 60° with the horizontal in the same direction.
The maximum distance between them till both of them
strike the ground is approximately g = 10 ms −2
(
)
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5.59
Chapter 5: Kinematics II
(A) 10 m
(C) 23 m
(B) 16 m
(D) 30 m
54. Two particles are projected simultaneously in the same
vertical plane from the same point, with different
speeds u1 and u2 , making angles θ1 and θ 2 respectively with the vertical, such that u1 sin θ1 = u2 sin θ 2 .
The path followed by one, as seen by the other (as long
as both are in flight), is
(A)a straight line making an angle θ1 − θ 2 with the
horizontal.
(B) a parabola.
(C) a horizontal straight line.
(D) a vertical straight line.
55. A ball rolls of the top of a stair way with a horizontal
velocity u ms −1 . If the steps are h m high and b m
wide, the ball will hit the edge of the nth step, if
(A) n =
2 hu
gb 2
2 hu2
(C) n =
gb
(B)
n=
2 hu2
gb 2
hu2
(D) n = 2
gb
56. The trajectory of a projectile in a vertical plane is
y = ax − bx 2 , where a , b are constants and x , y are
respectively the horizontal and vertical distances of
the projectile from the point of projection. The maximum height attained by the particle and the angle of
projection from the ­horizontal are
(A)
b2
, tan −1 ( 2a )
2a
(B)
a2
, tan −1 ( 2a )
b
(C)
a2
, tan −1 ( a )
4b
(D)
2
2a
, tan −1 ( a )
b
57. The velocity of a particle moving in the x -y plane is
given by
dy
dx
= 8π sin ( 2π t ) and
= 8π cos ( 2π t )
dt
dt
when t = 0 , x = 8 and y = 0 . The path of the ­particle
is
(A) a straight line
(B) a circle
(C) an ellipse
(D) a parabola
58. A projectile is fired with a velocity u at right angle
to the slope which is inclined at an angle θ with the
horizontal. The expression for R is
05_Kinematics 2_Part 2.indd 59
u
R
θ
(A)
2u2
tan θ
g
(B)
2u2
sec θ
g
(C)
u2
tan 2 θ
g
(D)
2u2
tan θ sec θ
g
59. The horizontal range and maximum height attained
by a projectile are R and H, respectively. If a cong
stant horizontal acceleration a =
is imparted to the
2
projectile due to wind, then its horizontal range and
maximum height will be
H⎞
⎛
⎜⎝ R + ⎟⎠ , H
2
H
(A) ( R + H ) ,
2
(B)
(C) ( R + 2 H ) , H
(D) ( R + H ) , H
60. A particle moves in space along the path z = ax 3 + by 2
dy
dx
=c=
, where a , b and c are
in such a way that
dt
dt
constants. The acceleration of the particle is
( 2ax 2 + 6by 2 ) kˆ
(C) ( bc 2 x + 2by ) kˆ
(A)
iˆ
(B)
(D)
iˆ
( 6 ac2x + 2bc2 ) kˆ
( 4bc2x + 3 ac2 ) kˆ
61. A stone is thrown from a point at a distance a from
a wall of height b . If it just clears the wall then the
maximum height h reached by the stone for angle of
projection α is
(A)
a 2 tan 2 α
4 ( a tan α − b )
(B)
a 2 sec 2 α
4 ( a sec α − b )
(C)
a 2 tan 2 α
4b
(D)
a 2 tan 2 α
4 ( a − b cot α )
62. An aeroplane flying at a constant speed releases a
food-packet for flood victims. As the packet drops
away from the aeroplane,
(A)it will always be vertically below the aeroplane
only if the aeroplane was flying at an angle of 45°
to the horizontal.
(B)it will always be vertically below the ­aeroplane
only if the aeroplane was flying horizontally.
(C)it will gradually fall behind the aeroplane if the
aeroplane was flying horizontally.
(D)it will always be vertically below the aeroplane.
11/28/2019 7:17:34 PM
5.60 JEE Advanced Physics: Mechanics – I
(C) a circle of radius a and centre at ( a, 0 )
(D) neither a circle nor an ellipse
64. Ratio of minimum kinetic energies of two projectiles
of same mass is 4 : 1 . The ratio of the maximum height
attained by them is also 9 : 1 . The ratio of their ranges
would be
(B) 8 : 1
(A) 16 : 1
(C) 6 : 1
(D) 2 : 1
65. A person at the point P aims his rifle at an angle of
20° with the horizontal so that the bullet fired is to hit
an object at A but the bullet hits at point B, a vertical
distance δ below A. If the initial velocity of the bullet
is 600 ms −1 and the point P is at a horizontal distance
of 1 km from the point A then,
(B) δ = 154 m
(A) δ = 1543 m
(C) δ = 154.3 m
(D) δ = 15.43 m
66. A particle moves along the positive branch of the curve
x2
1
with x governed by x = t 2 , where x and y
y=
2
2
are measured in metre and t in second. At t = 2 s , the
velocity of the particle is
(A) 2i − 4 j ms −1
(B)
(C) 4i + 2j ms −1
(D) 4i − 2j ms −1
(A) 5 ms −2
(B) 10 ms −2
(C) 15 ms −2
(D) 20 ms −2
68. A particle is moving in x-y plane. At certain instant
of time, the components of its velocity and acceleration are vx = 3 ms −1 , vy = 4 ms −1 , ax = 2 ms −2
and ay = 1 ms −2 . The rate of change of speed at this
moment is
5 ms −2
(C) 2 ms −2
(B)
10 ms −2
(D) 4 ms −2
69. A particle is projected at an angle of 60° above the
horizontal with a speed of 10 ms −1 . After some time
the direction of its velocity makes an angle of 30°
05_Kinematics 2_Part 2.indd 60
(A)
5
ms −1
3
(C) 5 ms −1
(B)
5 3 ms −1
(D)
10
ms −1
3
70. A particle is thrown with a speed u at an angle θ with
the vertical. When the particle makes an angle ϕ with
the vertical, its speed changes to v .
v = u sin θ cos ϕ
(A) v = u cos θ
(B)
⎛ sec ϕ ⎞
(C) v = u ⎜
⎝ sec θ ⎟⎠
⎛ cosec ϕ ⎞
(D) v = u ⎜
⎝ cosec θ ⎟⎠
71. At a height 0.4 m from the ground, the velocity of a
projectile in vector form is v = 6i + 2j ms −1 . The
(
(
)
angle of projection with the vertical is g = 10 ms −2
(A) 45°
(B)
(C) 30°
⎛ 3⎞
(D) tan −1 ⎜ ⎟
⎝ 4⎠
)
60°
72. A car moves round a turn of constant curvature
between A and B (curve AB = 100 metre ) with a
steady speed of 72 kmh −1 . If an accelerometer were
mounted in the car, the magnitude of acceleration it
would record between A and B is
2i + 4 j ms −1
67. A particle moves in the x -y plane according to the
equations x = 4t 2 + 4t + 5 and y = −t 3 + 12t + 3 . At
t = 1 s , the acceleration of the particle is
(A)
above the horizontal. The speed of the particle at this
instant is
A
B
63. Position vector of a particle moving in x-y plane at
time t is r = a ( 1 − cos ( ωt ) ) i + a sin ( ωt ) j . The path of
the particle is
(A) an ellipse
(B) a circle of radius a and centre at ( 0 , 0 )
100 m
45°
3.14 ms −2
(A) ZERO
(B)
(C) 31.4 ms −2
(D) 6.28 ms −2
73. After one second the velocity of a projectile makes an
angle of 45° with the horizontal. After another 3 s, it
is travelling horizontally. The magnitude of its initial
velocity and angle of ­projection are g = 10 ms −2
(
)
⎛ 3⎞
(A) 50 ms −1 , tan −1 ⎜ ⎟
⎝ 4⎠
(B)
⎛ 4⎞
25 ms −1 , tan −1 ⎜ ⎟
⎝ 3⎠
⎛ 4⎞
(C) 50 ms −1 , tan −1 ⎜ ⎟
⎝ 3⎠
⎛ 3⎞
(D) 25 ms −1 , tan −1 ⎜ ⎟
⎝ 4⎠
74. The minimum speed with which a particle must be
projected from the origin so that it just passes through
the point P ( 30 m , 40 m ) , taking g = 10 ms −2 , is
11/28/2019 7:17:48 PM
Chapter 5: Kinematics II
(A) 30 ms −1
(B)
40 ms −1
(A) 4.7 ms −1
(B)
(C) 50 ms −1
(D) 60 ms −1
(C) 14.7 ms −1
(D) 47 ms-1
75. Two shots are projected from a gun at the top of a cliff
with the same velocity u at angles of projection α and
β with the horizon respectively. If the shots strike the
horizontal ground through the foot of the cliff at the
same point and if h is the height of the cliff, then
(A) h =
(B)
h=
2
2u ⎛ 1 − tan α ⎞
g ⎜⎝ 1 − tan β ⎟⎠
2u
( tan α + tan β )
g
(D) h =
2u2
cot ( α + β )
g
76. An object is thrown horizontally from a tower at A
and hits the ground 3 second later at B . The line of
sight from A to B makes an angle of 30° with the
horizontal. The initial velocity of the object, taking
g = 10 ms −2 is
A
30°
V0
B
(A) 10 ms −1
(B) 13 ms −1
(C) 26 ms −1
(D) 28.8 ms −1
77. A point on the rim of a flywheel has a peripheral speed
of 10 ms −1 at an instant when it is decreasing at the
rate of 60 ms −2 . If the magnitude of the total acceleration of the point at this instant is 100 ms −2 , the radius
of the flywheel is
(A) 25 metre
(B) 12.5 metre
(C) 2.5 metre
(D) 1.25 metre
78. Water flows from a horizontal pipe which is fixed at a
height of 2 m from the ground. If it falls at a distance
of 3 m as shown in figure, the speed of water when it
leaves the pipe is
2m
3m
05_Kinematics 2_Part 2.indd 61
(A)
u
1 + 2 cos 2 θ
2
(B)
u
1 + cos 2 θ
2
(C)
u
1 + 3 cos 2 θ
2
(D) ucos θ
80. Shots are fired at the same instant from the top and
bottom of a vertical cliff at angle α and β and they
strike an object simultaneously at the same point. If
the horizontal distance of the object from the cliff is l
and h be the height of the cliff then
2
(C) h =
7.4 ms −1
79. A particle is projected from the ground with an initial
speed of u at an angle θ with horizontal. The average
velocity of the particle between its point of projection
and highest point of trajectory is
2
2u
( tan α − tan β )
g
5.61
(A) h = l ( cot β − cot α )
(B)
h = l ( cos β − cos α )
(C) h = l ( tan β − tan α )
(D) h = l ( sin β − sin α )
81. A ball thrown upward at an angle of 30° to the horizontal lands on the top edge of a building 20 metre
away. The top edge is 5 metre above the throwing
point. The ball was thrown with a velocity of
(A) 20 ms −1
(B)
30 ms −1
(C) 40 ms −1
(D) 60 ms −1
82. A particle is projected vertically upwards from O
with velocity v and a second particle is ­projected at
the same instant from P (at a height h above O ) with
velocity v at an angle of projection θ . The time when
the distance between them is minimum is
(A)
h
2v sin θ
(B)
h
2v cos θ
(C)
h
v
(D)
h
2v
83. A projectile is given an initial velocity of i + 2j . The
cartesian equation of its path is g = 10 ms −2
(
(A) y = 2x −
2
x
5
(C) y = x − 5x 2
(B)
)
y = 2x − x 2
(D) y = 2x − 5x 2
84. A ball is launched with an initial velocity of 20 2 ms −1
making at angle 45° with horizontal. The angular
velocity of the particle at highest point of its journey
about point of projection is
(A) 0.1 rads −1
(B)
0.2 rads −1
(C) 0.3 rads −1
(D) 0.4 rads −1
11/28/2019 7:17:58 PM
5.62 JEE Advanced Physics: Mechanics – I
85. An object is projected up the incline at the angle shown
in figure with an initial velocity of 30 ms −1 . The distance x up the incline at which the object lands is
y
A
30 ms–1
O
30°
30°
(A) 600 m
(C) 60 m
x
l
(B) 104 m
(D) 208 m
86. A very broad elevator is going up vertically with a
constant acceleration of 2 ms −2 . At the instant when
its velocity is 4 ms −1 a ball is projected from the floor
of the lift with a speed of 4 ms −1 relative to the floor
at an elevation of 30° . The time taken by the ball to
return the floor is g = 10 ms −2
(
)
(A) 1 s
(B)
1
s
2
1
s
3
(D)
1
s
4
(C)
(D) 2 3 m
3m
88. An aircraft moving with a speed of 1000 kmh −1 is at
a height of 6000 m , just overhead of an anti-aircraft
gun. If the muzzle velocity is 540 ms −1 , the firing
angle θ should be
1000 kmh–1
V0 = 540 ms–1
(B) 30°
(D) 45°
( 14, 4 ) m at t = 2 s , the equation of the path is
(C) x = y + 2
05_Kinematics 2_Part 2.indd 62
(A) 5 ms −1
(B) 10 ms −1
(C) 20 ms −1
(D) 40 ms −1
92. A boy throws a ball with a velocity u at an angle α
with the vertical. At the same instant he starts running
with uniform velocity to catch the ball before it hits the
ground. To achieve this, he should run with a velocity
of
(B) usin α
(A) ucos α
(C) u tan α
(D)
u2 tan α
(B)
(A) 11 × 10 5 ms −2
(B) 1.1 × 10 5 ms −2
(C) 2.2 × 10 4 ms −2
(D) 2.2 × 10 5 ms −2
94. A projectile is thrown with an initial velocity of
ai + b j ms −1 . If the range of the projectile is twice the
89. A particle moves in the x -y plane with velocity
vx = 8t − 2 and vy = 2 . If it passes through the point
2
91. A particle is projected from the ground with an ­initial
velocity of 30 ms −1 at an angle of 60° with ­horizontal.
The magnitude of change in velocity in 2 s is
acceleration at the edge of the rotor is ( take π 2 = 10 )
θ
(A) x = y 2 − y + 2
(A) the other end also moves uniformly
(B) the speed of other end goes on decreasing
(C) the speed of other end goes on increasing
(D)the speed of other end first decreases and then
increases
93. The rotor of a turbine rotates at the rate of 2000 rpm .
If the diameter of the rotor is 5 m , the centripetal
6000 m
(A) 73°
(C) 59°
x
( g = 10 ms−2 )
87. In PROBLEM 86, range of the ball over the floor of the
lift is
2
(A) 2 m
(B)
m
3
(C)
90. A rod of length l leans by its upper end against a
smooth vertical wall, while its other end leans against
the floor. The end that leans against the wall moves
uniformly downward with velocity v0 . Then
x=y+2
2
(D) x = y + y − 2
(
)
maximum height reached by it, then
(A) a = 2b
(B)
b=a
(C) b = 2 a
(D) b = 4 a
95. Time taken by the projectile to reach from A to B is t .
Thet the distance AB is equal to
11/28/2019 7:18:08 PM
Chapter 5: Kinematics II
u
A
(A) 6 s
(C) 1.5 s
B
60°
30°
3ut
2
(A)
ut
3
(B)
(C)
3ut
(D) 2ut
96. A stone is projected from a horizontal plane. It attains
maximum height H and strikes a stationary smooth
wall and falls on the ground ­vertically below the maximum height. Assume the collision to be elastic, the
height of the point on the wall where stone will strike is
(A)
H
2
(B)
H
4
(C)
3H
4
(D)
H
3
97. A circular disc of radius r = 5 m is rotating in horizontal plane about y-axis. Y-axis is vertical axis passing
through the centre of disc and x -z is the horizontal
plane at ground. The height of disc above ground is
h = 5 m . Small particles are ejecting from disc in horizontal direction with speed 12 ms −1 from the circumference of disc then the distance of these particles from
origin when they hits the x -z plane is
(A) 12 m
(B) 13 m
(C) 5 m
(D) 6 m
98. It was calculated that a shell when fired from a gun
5π
with a certain velocity and at an angle of elevation
36
radians should strike a given target. In actual practice
it was found that a hill just prevented in the trajectory.
At what angle of elevation should the gun be fired to
hit the target.
5π
11π
radians
(B)
radians
(A)
36
36
(C)
7π
radians
36
(D)
13π
radians
36
99. A body is thrown with speed of 30 ms −1 at angle 30°
with horizontal from a perfectly inelastic horizontal
floor. The time after which it is moving perpendicular
to its initial direction of motion is
05_Kinematics 2_Part 2.indd 63
5.63
(B) 3 s
(D) never
100. A body is thrown from a point with speed 50 ms −1
at an angle 37° with horizontal. When it has moved
a horizontal distance of 80 m then its distance from
point of projection is
(A) 40 m
(B)
(C) 40 5 m
(D) None of these
40 2 m
101. A ball is thrown from ground level so as to just clear
a wall 4 meters high at a distance of 4 meters and
falls at a distance of 14 meters from the wall, then the
magnitude of the velocity of the ball is
(A)
281 ms −1
(B)
812 ms −1
(C)
182 ms −1
(D)
128 ms −1
102. An object is thrown at an angle α to the horizontal
( 0° < α < 90° ) with a velocity. Then ­during ascent
(ignoring air drag) the acceleration
(A)with which the object moves is g , at all points.
(B) tangential to the path decreases.
(C)normal to the path increases, becoming equal to
g at the highest point.
(D) All of these
103. A projectile is thrown with a velocity of 20 ms −1 , at
an angle of 60° with the horizontal. After how much
time the velocity vector will make an angle of 45°
(
with the horizontal? Take g = 10 ms −2
(A)
(C)
3s
( 3 + 1) s
)
(B)
1
s
3
(D)
( 3 − 1) s
104. A golfer standing on level ground hits a ball with a
velocity of u = 52 ms −1 at an angle θ above the hori5
zontal. If tan θ =
, then the time for which the ball
12
is at least 15 m above the ground (i.e. between A
(
and B ) will be take g = 10 ms −2
u
θ
(A) 1 s
(C) 3 s
A
)
B
15 m
15 m
(B) 2 s
(D) 4 s
105. Which of the following ideas is helpful in
­understanding projectile motion?
11/28/2019 7:18:16 PM
5.64 JEE Advanced Physics: Mechanics – I
(A) vx2 + vy2 = constant
(B)Acceleration is + g when the object is rising and
−g when falling
(C)
In the absence of friction the trajectory will
depend on the object’s mass as well as its initial
velocity and launch angle
(D)
The horizontal motion is independent of the
­vertical motion
111. A projectile is thrown at angle b with vertical. It
reaches a maximum height H . The time taken to
reach highest point of its path is
(A)
H
g
(B)
2H
g
(C)
H
2g
(D)
2H
g cos β
106. A ball is projected upwards from the top of the tower
with a velocity 50 ms −1 making an angle 30° with
the horizontal. The height of tower is 70 m . After
how many seconds the ball will strike the ground?
(A) 3 s
(B) 5 s
(C) 7 s
(D) 9 s
112. A body is projected at angle 45° to horizontal with
velocity 20 ms −1 from ground. If there is an acceleration in horizontal direction of 2 ms −2 , then calculate
horizontal range of this particle
(A) 40 m
(B) 48 m
(C) 8 m
(D) 20 m
g 2
x . The
2
113. A particle is projected with a velocity of 20 ms −1 at
an angle of 30° to an inclined plane of inclination 30°
to the horizontal. The particle hits the inclined plane
at an angle 30°, during its journey. The time of flight is
107. The equation of projectile is y = 3 x −
angle of projection and initial velocity is
(A) 30° , 4 ms −1
(B)
60° , 2 ms −1
(C) 60° , 4 ms −1
(D) 90° , 4 ms −1
108. A projectile is fired at an angle of 30° to the horizontal such that the vertical component of its initial
velocity is 80 ms −1 . Its time of flight is T . Its velocity
T
has a magnitude of nearly
at t =
4
(A) 200 ms −1
(C) 140 ms −1
(B) 300 ms −1
(D) 100 ms −1
109. A particle is projected from a point A with velocity
u 2 at an angle of 45° with horizontal as shown in
figure. It strikes the plane BC at right angles. The
velocity of the particle at the time of collision is:
u 2
C
(A)
20 sin ( 60° )
g
(B)
20 sin ( 60° )
g cos ( 30° )
(C)
20 sin ( 30° )
g cos ( 60° )
(D)
20 sin ( 30° )
g
114. A particle starts flying in the xy-plane with a speed of
2iˆ + 5xjˆ . Initial position of the particle was the origin
( 0, 0 ) of the plane. The trajectory of the particle is
represented by the equation
(A) y = 1.25x 2
(B)
y = 5x 2
(C) y = 2.5x 2
(D) x = 5 y 2
115. The ceiling of a tunnel is 5 m high. What is the maximum horizontal distance that a ball thrown with a
speed of 20 ms −1 , can go without hitting the ceiling
(A)
(C)
3u
2
2u
3
(
60°
45°
A
B
(B)
u
2
(D) u
110. Consider a boy on a trolley who throws a ball with
speed 20 ms −1 at an angle 37° with respect to trolley in direction of motion of trolley which moves
horizontally with speed 10 ms −1 then what will be
maximum distance travelled by ball parallel to road
(A) 20.2 m
(B) 12 m
(C) 31.2 m
(D) 62.4 m
05_Kinematics 2_Part 2.indd 64
of the tunnel? Take g = 10 ms −2
)
(A) 30 m
(B)
40 m
(C) 30 2 m
(D) 20 3 m
116. In projectile motion, the modulus of rate of change of
speed
(A) is constant
(B) first increases then decreases
(C) first decreases then increases
(D) None of these
117. If a stone is to hit at a point which is at a horizontal
distance d away and at a height h above the point
from where the stone starts, then what is the value of
initial speed u if the stone is launched at an angle q?
11/28/2019 7:18:26 PM
5.65
Chapter 5: Kinematics II
are in metre and t in second. The acceleration of the
particle at t = 5 s is
u
h
(A) 40 ms −2
θ
(C) 8 ms
d
(A)
(B)
(C)
(D)
g
cos θ
d
2 ( d tan θ − h )
d
cos θ
2 ( d tan θ − h )
0
gd 2
h cos 2 θ
2
gd
d−h
θ
119. A large box is moving on horizontal floor with constant acceleration a = g . A particle is projected inside
box with velocity u and angle θ with horizontal
with respect to box frame. For the given u , the value
of θ for which horizontal range inside box will be
maximum is
π
π
(B)
(A)
4
8
(D)
π
3
120. An object is projected with a velocity of 20 ms −1 making an angle of 45° with horizontal. The equation for
the trajectory is h = Ax − Bx 2 where h is height, x
is horizontal distance. A and B are constants. The
ratio A : B is g = 10 ms −2
(A) 1 : 5
(C) 1 : 40
ωx ⎞
(C) y = a sin ⎛
⎜⎝ v ⎟⎠
0
ωx ⎞
(D) y = a cos ⎛
⎜⎝ v ⎟⎠
0
(B) 5 : 1
(D) 40 : 1
y = a cos ( ωt )
123. Ratio of minimum kinetic energies of two projectiles of same mass is 4 : 1 . The ratio of the maximum
height attained by them is also 4 : 1 . The ratio of their
ranges would be
(A) 16 : 1
(B) 4 : 1
(C) 8 : 1
(D) 2 : 1
124. At a height 0.4 m from the ground, the velocity of a
projectile is, v = 6iˆ + 2 ˆj ms −1 . The angle of projec-
(
−2
)
)
(A) 45°
(B)
(C) 30°
⎛ 3⎞
(D) tan −1 ⎜ ⎟
⎝ 4⎠
60°
125. A person standing on a truck moving with a uniform
velocity 14.7 ms −1 on a horizontal road throws a ball
in such a way that it returns to him after 4 s. The
speed and angle of projection as seen by a man on the
road are
(A) 19.6 ms −1 , vertical
(B)
24.5 ms −1 , vertical
(C) 19.6 ms −1 , 53° with the road
(D) 24.5 ms −1 , 53° with the road
126. Trajectories of two projectiles are shown in the figure.
Let T1 and T2 be the time periods and u1 and u2 be
their speeds of projection. Then
Y
)
121. The x and y coordinates of a particle at any time t
are given by x = 2t + 4t 2 and y = 5t , where x and y
05_Kinematics 2_Part 2.indd 65
(B)
(
θ
3π
8
(A) y = a sin ( ωt )
tion is g = 10 ms
(A)The particles will collide the plane with same
speed
(B) The times of flight of each particle are same
(C)Both particles strikes the plane perpendicularly
(D) None of these
(
(D) ZERO
The equation of trajectory of the particle is
118. From an inclined plane two particles are projected
with same speed at same angle θ , one up and other
down the plane as shown in figure. Which of the following statement(s) is/are correct?
(C)
20 ms −2
122. A particle starts from the origin at t = 0 . It moves in a
plane with a velocity given by v = v iˆ + ( aω cos ωt ) ˆj .
g
θ
(B)
−2
1
2
X
(A) T2 > T1
(B)
T1 > T2
(C) u1 > u2
(D) u1 < u2
11/28/2019 7:18:36 PM
5.66 JEE Advanced Physics: Mechanics – I
Multiple Correct Choice Type Questions
This section contains Multiple Correct Choice Type Questions. Each question has four choices (A), (B), (C) and (D), out of
which ONE OR MORE is/are correct.
1.
A bead is free to slide down a smooth wire tightly
stretched between the points P1 and P2 on a vertical
circle of radius R. If the bead starts from rest from P1,
the highest point on the circle and P2 lies anywhere on
the circumference of the circle. Then,
P1
g
θ
P2
4.
R
5.
(A)time taken by bead to go from P1 to P2 is dependent on position of P2 and equals 2
R
cos θ .
g
(B)time taken by bead to go from P1 to P2 is independent of position of P2 and equals 2
R
.
g
2.
⎛ u⎞
(D) time of flight from P to Q is ⎜ ⎟ c osecθ
⎝ g⎠
05_Kinematics 2_Part 2.indd 66
(B)
v = −2j
(
)
(D) r = 4.5i − 2.25j m
Two particles projected from the same point with same
speed u at angles of projection α and β strike the
horizontal ground at the same point. If h1 and h2 are
the maximum heights attained by projectiles, R be the
range for both and t1 and t2 be their time of flights
respectively then
From a point P , a particle is projected with a velocity
u at an angle θ with horizontal. At a certain point Q ,
the particle moves at right angles to its initial direction
of motion. Then
(A) velocity of particle at Q is usin θ .
(C) speed of particle at Q is ucot θ .
A particle is launched from the origin with an i­nitial
velocity u = ( 3i ) ms −1 under the influence of a con1 ⎞
⎛
stant acceleration a = − ⎜ iˆ + ˆj ⎟ ms −2 . Its velocity v
⎝
2 ⎠
and position vector r when it reaches its maximum
x-co-ordinate are
7.
T⎞
⎛ T⎞⎛
y = 4H ⎜ ⎟ ⎜ 1 − ⎟
⎝ t ⎠⎝
t⎠
⎛ u⎞
(B) time of flight from P to Q is ⎜ ⎟ sec θ
⎝ g⎠
h
.
g
A projectile is projected from the ground making an
angle α with the horizontal. Air exerts a drag which
is proportional to the velocity of the projectile
(A)the time of ascent will be equal to the time of
descent
(B)the time of ascent will be greater than time of
descent
(C)the time of descent will be greater than time of
ascent
(D) at highest point velocity will be horizontal
R⎞
⎛ R⎞⎛
(D) y = 4 H ⎜ ⎟ ⎜ 1 − ⎟
⎝ x⎠⎝
x⎠
3.
(D) time of passing between the walls is 2
6.
t⎞
⎛ t ⎞⎛
(A) y = 4 H ⎜ ⎟ ⎜ 1 − ⎟
⎝ T⎠⎝
T⎠
x⎞
⎛ x⎞⎛
(C) y = 4 H ⎜ ⎟ ⎜ 1 − ⎟
⎝ R⎠⎝
R⎠
2h
.
g
( )
(C) r = ( 3i − 2j ) m
For an oblique projectile, if T is the total time of flight,
H the maximum height and R is the horizontal
range, then x and y co-ordinates at any time t are
related as (neglect air drag)
(B)
(C) time of passing between the walls is
(A) v = −1.5j ms −1
(C)acceleration of bead along the wire is g cos θ .
(D)velocity of bead when it arrives at P2 is 2 gR cos θ .
A particle is projected with a velocity 2 hg so that
it just clears two walls of equal height h at horizontal
separation 2h from each other. Then the
(A) angle of projection is 30° with vertical.
(B) angle of projection is 30° with horizon.
(A) α + β =
(C)
8.
π
2
t1
= tan α
t2
(B)
R = 4 h1h2
(D) tan α =
h1
h2
The co-ordinates of a particle moving in a plane are
given by x = a cos ( pt ) and y = b sin ( pt ) , where a ,
b ( < a ) and p are positive constants of appropriate
dimensions. Then
11/28/2019 7:18:46 PM
Chapter 5: Kinematics II
(A) the path of the particle is an ellipse.
(B)the velocity and acceleration of the particle are
π
.
normal to each other at t =
2p
(C)the acceleration of the particle is always directed
towards the focus.
(D)
the distance travelled by the particle in time
π
is a .
­interval t = 0 to t =
2p
9.
A ball starts falling freely from a height h from a point
on the inclined plane forming an angle α with the
horizontal as shown. After collision with the incline it
rebounds elastically off the inclined plane. Then
P
h
α
A
strikes the incline at A .
(B)
it again strikes the incline at t =
8h
after it
g
2h
after it
g
strikes the incline at A .
(C)it again strikes the incline at a distance 4h sin α
from A along the incline.
(D)it again strikes the incline at a distance 8h sin α
from A along the incline.
10. A cart is moving along +x direction with a velocity
of 4 ms −1 . A person on the cart throws a stone with a
velocity of 6 ms −1 with respect to himself. In the frame
of reference of the cart the stone is thrown in the y -z
plane making an angle of 30° with the vertical z-axis.
Then with respect to an observer on the ground
(A) the initial velocity of the stone is 10 ms −1 .
(B) the initial velocity of the stone is 2 13 ms −1 .
(C)the velocity at the highest point of its motion is
zero.
(D)the velocity at the highest point of its motion is
5 ms −1 .
11. Two shells are fired from a cannon successively with
speed u each at angles of projection α and β , respectively. If the time interval between the firing of shells is
t and they collide in mid air after a time T from the
firing of the first shell. Then
(A) T cos α = ( T − t ) cos β
(B) α > β
05_Kinematics 2_Part 2.indd 67
(C) ( T − t ) cos α = t cos β
(D) ( u sin α ) T −
1 2
1
2
gT = ( u sin β ) ( T − t ) − g ( T − t )
2
2
12. Choose the correct alternative (s)
(A)A ball is thrown vertically up and another ball is
thrown at an angle θ with the vertical such that
both of them remain in air for the same period of
time, then the ratio of heights attained by the two
balls is 1 : 1
(B)The time of flight T and the horizontal range
R of a projectile are connected by the equation
gT 2 = 2R tan θ , where θ is the angle of projection
(C)
For a projectile whose range R is n times its
maximum height H the angle of projection is
⎛ 4⎞
tan −1 ⎜ ⎟
⎝ n⎠
PA = h
(A)
it again strikes the incline at t =
5.67
(D)If the greatest height to which a man can throw a
stone is h , then the greatest horizontal distance
upto which he can throw the stone is 2h
13. A radar observer on the ground is watching an
approaching projectile. At a certain instant he has the
following information.
(i) The projectile has reached the maximum altitude
and is moving with a horizontal velocity v;
(ii) The straight line distance of the observer to the
projectile is l;
(iii) The line of sight to the projectile is at an angle θ
above the horizontal.
Assuming earth to be flat and the observer lying in the
plane of the projectile’s trajectory then,
(A)the distance between the observer and the point
v 2 sin θ
− l cos θ .
of impact of the projectile is D =
g
(B)the distance between the observer and the point of
impact of the projectile is D = v
2l sin θ
− l cos θ .
g
(C)the projectile will pass over the observer’s head
2v 2 tan θ sec θ
.
for l <
g
(D)the projectile will pass over the observer’s head
2v 2 tan θ sec θ
.
for l >
g
14. A projectile has the same range R for two angles of
projections. If T1 and T2 be the times of flight in the
two cases and θ be the angle of projection corresponding to the time T1 , then
11/28/2019 7:18:55 PM
5.68 JEE Advanced Physics: Mechanics – I
(A) T1T2 ∝ R2
(C)
T1
= tan θ
T2
(B)
T1
= cot θ
T2
(D) T1T2 ∝ R
15. Two guns situated at the top of a hill of height 10 m ,
fire one shot each with the same speed of 5 3 ms–1
at some interval of time. One gun fires horizontally
and other fires upwards at an angle of 60° with the
horizontal. The shots collide in mid air at the point P .
Taking the origin of the coordinate system at the foot
of the hill right below the muzzle, trajectories in x -y
plane and g = 10 ms −2 , then
(A)the first shell reaches the point P at t1 = 1 s from
the start.
(B)the second shell reaches the point P at t2 = 2 s
from the start.
(C)the first shell is fired 1 s after the firing of the second shell.
(D)they collide at P whose coordinates are given by
5 3, 5 m .
(
1
dR
=−
(A)
R
291
dR
1 ⎞
⎛ 1
=g
−
⎜
R
⎝ g p g e ⎟⎠
(B)
1
dR
=
R 291
(D)
dR
1 ⎞
⎛ 1
= −g
−
⎜
R
⎝ g p g e ⎟⎠
17. A boat is moving directly away from a cannon on the
shore with a speed v1 . The cannon fires a shell with
a speed v2 at an angle α and the shell hits the boat.
Then,
(A)the shell hits the boat when the time equal to
2v2 sin α
is lapsed.
g
2v1v2 sin α
(B)the boat travels a distance
from its
g
original position.
(C)the distance of the boat from the cannon at the
2
instant the shell is fired is ( v2 sin α ) ( v2 cos α − v1 ) .
g
(D)the distance of the boat from the cannon when the
2
shell hits the boat is ( v2 sin α )( v2 cos α ) .
g
18. A particle is projected from the ground with a v
­ elocity
40 2 ms −1 which makes an angle of 45° with the
horizontal. At time t = 2 s
05_Kinematics 2_Part 2.indd 68
(C)velocity makes an angle of tan −1 ( 2 ) with the
horizontal
(D) particle is at height of 60 m from ground
19. A particle P lying on smooth horizontal x -y plane
(
)
starts from 6iˆ + 8 ˆj m with velocity ( 2î ) ms −1 .
Another particle Q is projected (horizontally from
(
)
origin with velocity aiˆ + bjˆ so that is strikes P after
2 s . Then
(B) a = 5
(A) a = 2
(C) b = 4
(D) b = 8
20. A projectile is fired upward with velocity v0 at an
angle θ and strikes a point P ( x , y ) on the roof of the
building (as shown). Then,
y
Roof
)
16. A projectile is thrown with an initial velocity u , at an
angle of projection θ first from the equator and then
from the pole. The fractional decrement in the range of
projectile is
(C)
(A) displacement of particle is 100 m
(B) vertical component of velocity is 20 ms −1
v0
h
P(x, y)
α
θ
x
(A)the projectile hits the roof in minimum time if
π
θ +α = .
2
(B)the projectile hits the roof in minimum time if
π
θ +α = .
4
(C)the minimum time taken by the projectile to hit
the roof is tmin =
(D)
the
projectile
v0 − v02 − 2 gh cos 2 α
g cos α
never
reaches
the
.
roof
for
v0 < 2 gh cos α .
21. A particle is fired from a point on the ground with
speed u making an angle θ with the horizontal. Then
(A)the radius of curvature of the projectile at the
u2 cos 2 θ
highest point is
g
(B)the radius of curvature of the projectile at the
u2
highest point of launch is
g cos θ
(C)at the point of projection tangential acceleration is
g sin θ
(D)at the highest point tangential acceleration is zero
11/28/2019 7:19:04 PM
Chapter 5: Kinematics II
22. A shot is fired with a velocity u at an angle ( α + θ )
with the horizon from the foot of an incline plane of
angle α through the point of projection. If it hits the
plane horizontally then
(A) tan θ =
tan α
1 + 2 tan 2 α
(B)
(C) tan θ =
2 tan α
1 + 2 tan 2 α
(D) tan θ =
25. Path of a particle moving in x-y plane is y = 3 x + 4 .
At some instant suppose x-component of velocity is
1 ms −1 and it is increasing at a rate of 1 ms −2 . Then
at this instant
(A) speed of particle is 10 ms −1
(B) acceleration of particle is 10 ms −2
(C) velocity-time graph is a straight line
(D) acceleration-time graph is a straight line
tan θ = 2 tan α
sin α cos α
1 + sin 2 α
23. Two second after projection, a projectile is travelling
in a direction inclined at 30° to the horizon. After one
more second it is travelling horizontally. Then
(A) the velocity of projection is 20 ms −1 .
(B) the velocity of projection is 20 3 ms −1 .
(C) the angle of projection is 30° with vertical.
(D) the angle of projection is 30° with horizon.
24. A particle is projected from point A with speed u and
angle of projection is 60° . At some instant, magnitude
of velocity of particle is v and it makes an angle θ
with horizontal. If radius of curvature of path of parti8
cle at the given instant is
times minimum radius
3 3
of curvature during the whole flight, then
(A) θ = 37°
(B) θ = 30°
u
(C) v =
3
u
(D) v =
5.69
26. The co-ordinate of the particle in x -y plane are given
as x = 2 + 2t + 4t 2 and y = 4t + 8t 2 . The motion of the
particle is
(A) along a straight line
(B) uniformly accelerated
(C) along a parabolic path
(D) non-uniformly accelerated
27. Velocity of a particle moving in a curvilinear path
­varies with time as v = 2tiˆ + t 2 ˆj ms −1 , where, t is in
second. At t = 1 s , the
(
)
(A) acceleration of particle is 8 ms −2 .
(B)tangential acceleration of particle is
(C) radial acceleration of particle is
(D) None of these
6
ms −2 .
5
2
ms −2 .
5
2 3
Reasoning Based Questions
This section contains Reasoning type questions, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is
correct. Each question contains STATEMENT 1 and STATEMENT 2. You have to mark your answer as
Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.
Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1.
Bubble (C) If STATEMENT 1 is TRUE and STATEMENT 2 is FALSE.
Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE.
1.
Statement-1: For an oblique projectile launched from
the ground at an angle θ with the horizontal, R = H
at θ = tan −1 ( 4) .
Statement-2: Maximum range of projectile is proportional to square of initial velocity and inversely proportional to g .
2.
Statement-1: The net acceleration of a particle in circular motion is always directed radially inwards.
Statement-2: Whenever a particle moves in a circular
path, an acceleration exists which is directed towards
the centre.
05_Kinematics 2_Part 2.indd 69
3.
Statement-1: In uniform circular motion, acceleration
is constant.
Statement-2: In uniform circular motion, speed is
constant.
4.
Statement-1: When a particle is thrown obliquely from
the surface of the Earth, it always moves in a parabolic
path, provided the air resistance is negligible.
Statement-2: A projectile motion is a two-dimensional
motion.
5.
Statement-1: In uniform circular motion acceleration
is constant.
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5.70 JEE Advanced Physics: Mechanics – I
Statement-2: In uniform circular motion ­magnitude of
v2
acceleration is
and direction is always towards the
r
centre.
6.
Statement-2: Velocity along horizontal direction
remains same but velocity along vertical direction is
changed. When particle strikes the ground, magnitude
of final vertical velocity is equal to magnitude of initial vertical velocity.
7.
10. Statement-1: Two particles of different mass, projected
with same velocity at same angles. The maximum
height attained by both the particles will be same.
Statement-2: The maximum height of projectile is
independent of the mass of the particle.
Statement-1: A particle is moving on a horizontal surface. Its path will be straight line if initial velocity and
acceleration are collinear and/or either of them is zero.
Statement-2: Angle between u and a determine path
therefore, it may be accelerated or retarded curvilinear.
Statement-1: When a particle moves in a circle with
a uniform speed, its velocity and acceleration both
changes.
Statement-2: The centripetal acceleration in circular
motion is dependent on angular velocity of the body.
8.
Statement-1: A coin is allowed to fall in a train moving
with constant velocity. Its trajectory is parabola as seen
by an observer sitting in the train.
Statement-2: An observer on ground will see the path
of coin as a parabola.
9.
Statement-1: A particle is projected with a velocity u
making at an angle θ < 90° with the horizontal. When
particle strikes the ground its speed is again u .
11. Statement-1: When speed of projection of a body is
made n -times, its time of flight becomes n times
Statement-2: This is because range of projectile become
n times.
12. Statement-1: Horizontal velocity of a particle moving
under the influence of gravity remains constant.
Statement-2: Acceleration due to gravity acts vertically downwards.
13. Statement-1: An oblique projectile is launched from
the ground to attain a maximum range. The maximum
height attained by the projectile is 25% of range.
Statement-2: R =
u2 sin ( 2θ )
u2 sin 2 θ
and H =
2g
g
Linked Comprehension Type Questions
This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph
followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For the sake of
competitiveness there may be a few questions that may have more than one correct options)
Comprehension 1
(A) ( 2kα ) ˆj
(B)
A point moves in the plane x , y according to the law
x = kt , y = kt ( 1 − α t ) , where k and α are positive constants and t is the time. Based on the above facts, answer
the following questions.
kα ⎞ ˆ
(C) ⎛⎜
j
⎝ 2 ⎟⎠
(D) −2 ( kα ) ˆj
1.
The trajectory y ( x ) followed by the particle is a/an
(A) straight line
(C) parabola
2.
The velocity v of the point is minimum at time
(A) t =
1
α
2
(C) t =
α
3.
(B) ellipse
(D) cycloid
(B)
t=
1
2α
1
(D) t =
3α
The acceleration of the particle at time when the velocity has a minimum value is
05_Kinematics 2_Part 2.indd 70
4.
( kα ) ˆj
The moment t0 at which the velocity vector forms an
π
angle
with the acceleration vector is
4
(A)
1
α
(B)
1
2α
(C)
2
α
(D)
1
3α
Comprehension 2
A particle is projected from a point A with velocity u 2
at an angle of 45° with horizontal as shown in figure. It
strikes the plane BC at right angles. Based on the above
facts, answer the following questions.
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Chapter 5: Kinematics II
C
P
u 2
B
(D)
(C)
10. The minimum value of horizontal velocity of particle
at point P so that particle may strike the plane AB is
3u
g
u
g
(B)
u ⎛ 3 + 1⎞
⎟
g ⎜⎝
3 ⎠
⎛
3 ⎞u
(D) ⎜
⎝ 3 − 1 ⎟⎠ g
(
9.
)
The launch speed in ms −1 is
(A) 20 ms −1
(B)
(C) 40 ms −1
(D) 40 2 ms −1
20 2 ms −1
The value of θ (in radian) is
π
(B)
(A)
6
(C)
π
3
π
4
2π
(D)
3
The coordinate, where the maximum height is attained
is
( 80, 20 ) m
(C) ( 20 , 80 ) m
(A)
( 20, 40 ) m
(D) ( 40 , 20 ) m
(B)
Comprehension 4
Consider the situation shown in the figure. A particle has
to be projected from point P in horizontal direction. It is
required for the particle to strike the plane AB (see figure).
05_Kinematics 2_Part 2.indd 71
B
2a
An oblique projectile is launched from a point O with an
initial velocity u that makes an angle θ with the horizontal. It is observed that the projectile attains a maximum
R
height H at the point ⎛⎜ , H ⎞⎟ , where R is the maximum
⎝ 2
⎠
distance from O at which the projectile strikes the ground.
Now, if the velocity of the projectile is v = 20iˆ + 10 ˆj ms −1
when it is at a height of 15 m above the ground, then based
on this information and taking g = 10 ms −2 , answer the following questions.
8.
a
A
Comprehension 3
7.
a
a
2u
3
The time after which collision takes place is
(A)
a
a
The velocity of the particle at the time of collision is
u
(A) u
(B)
2
(C)
6.
a
60°
A
3u
2
u
a
45°
5.
5.71
(A)
9
ag
8
(B)
25
ag
8
(C)
1
ag
2
(D)
3ag
11. The time taken by particle in going from point P to
plane AB is
(A) 4
a
g
(B)
4a
g
(C)
2a
g
(D)
8a
g
12. The maximum value of horizontal velocity of particle
at point P so that it may strike the plane AB is
(A)
9
ag
8
(B)
25
ag
8
(C)
1
ag
2
(D)
3ag
Comprehension 5
A ball is projected horizontally from a height of 100 m from
the ground with a speed of 20 ms −1 . Taking g = 10 ms −2
and based on the information provided, answer the following questions.
13. The time taken to reach the ground is
(A)
5s
(C) 3 5 s
(B)
2 5s
(D) 4 5 s
14. The horizontal distance it covers before striking the
ground is
(A) 20 5 m
(B)
(C) 60 5 m
(D) 80 5 m
40 5 m
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5.72 JEE Advanced Physics: Mechanics – I
15. The angle with the vertical which it strikes the ground
is
⎛ 1 ⎞
tan −1 ⎜
⎝ 2 ⎟⎠
⎛ 1 ⎞
(A) tan −1 ⎜
⎝ 3 ⎟⎠
(B)
⎛ 1 ⎞
(C) tan −1 ⎜
⎝ 5 ⎟⎠
⎛ 1 ⎞
(D) tan −1 ⎜
⎝ 7 ⎟⎠
22. The horizontal range of the projectile is
(A) 3.2 m
(B) 4.9 m
(C) 8.7 m
(D) 9.6 m
Comprehension 7
Comprehension 6
For a particle moving in the x -y plane the x , y coordinates
as a function of time are given by x = 6t and y = 8t − 5t 2 ,
where x and y are in metre and t is in second. Assume
no air drag, answer the following questions. Based on the
above facts, answer the following questions.
16. Select the correct statement from the following
(A)The particle will follow a parabolic path to have
a projectile motion with an initial velocity of
−1
−2
5 ms
and an acceleration of 3 ms
acting
along −y axis
(B)The particle will follow a parabolic path to have
a projectile motion with an initial velocity of
6 ms −1 and a constant acceleration of 5 ms −2
acting along −y axis
(C)The particle will follow a parabolic path to have
a projectile motion with an initial horizontal
velocity of 6 ms −1 and a constant acceleration of
10 ms
−2
21. The maximum height attained by the projectile is
(A) 0.8 m
(B) 1.6 m
(C) 2.9 m
(D) 3.2 m
The maximum height attained by an oblique projectile is
8 m and the horizontal range is 24 m. Based on the above
facts, answer the following questions.
23. The vertical component of the velocity of projection is
(A)
24. The horizontal component of the velocity of ­projection
is
(B)
< 6 ms −1
(C) 6 ms −1
(D) zero
18. The initial vertical velocity is
(B)
(C) 4 g
(D) 5 g
3 g
(A) 2 g
(B)
(C) 4 g
(D) 5 g
3 g
26. The angle of projection is
(A) cos −1 ( 0.8 )
(C) tan
−1 (
0.6 )
(B)
sin −1 ( 0.8 )
(D) cot −1 ( 0.8 )
Comprehension 8
A particle initially at rest and starting from the origin
is moving under the influence of acceleration given by
a = 6tiˆ + 8tjˆ ms −2 . Based on the above facts, answer the
(
)
following questions.
(A) 6 ms −1
(B)
(C) 8 ms −1
(D) 10 ms −1
7 ms −1
19. The velocity of projection of the projectile is
(A) 6 ms −1
(B)
(C) 8 ms −1
(D) 10 ms −1
7 ms −1
20. The time of ascent of the projectile is
(A) 0.2 s
(B) 0.4 s
(C) 0.6 s
(D) 0.8 s
05_Kinematics 2_Part 2.indd 72
(A) 2 g
25. The velocity of projection is
(D)None of the above statement seems to be correct
(A) > 6 ms −1
2 g
(D) 4 g
(C) 3 g
acting along −y axis
17. The velocity, of the projectile, along x-axis after
0.2 ­second is
(B)
g
27. Velocity of particle at t = 3 s
(A) 45 ms −1
(B)
40 ms −1
(C) 35 ms −1
(D) 22 ms −1
28. Displacement of particle at t = 3 s is
(A) 28 m
(B) 30 m
(C) 35 m
(D) 45 m
29. Path of particle will be
(A) Straight line
(C) Circle
(B) Parabola
(D) None of these
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Chapter 5: Kinematics II
5.73
Matrix Match/Column Match Type Questions
Each question in this section contains statements given in two columns, which have to be matched. The statements in
COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are labelled p, q, r, s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples:
A
B
C
D
p
p
p
p
p
q
q
q
q
q
r
s
t
r
r
r
r
s
s
s
s
t
t
t
t
If the correct matches are A → p, s and t; B → q and r; C → p and q; and D → s and t; then the correct darkening of bubbles
will look like the following:
1.
A particle is launched with an initial velocity of
3.
20 2 ms −1 making an angle of 45° with the horizontal. Based on this information and g = 10 ms −2 match
the contents of COLUMN-I with their counterparts in
COLUMN-II.
A particle is moving in a curvilinear path such that its
velocity v , in ms −1 , at any instant of time t , in sec
ond, is given by v = 2tiˆ + t 2 ˆj . Match the quantities in
COLUMN-I calculated at t = 1 s, with the respective
values in COLUMN-II.
COLUMN-I
COLUMN-II
COLUMN-I
COLUMN-II
(A) Magnitude of average velocity, in
(p) 25 5
(A) Tangential acceleration, in
ms −2
(p) 2 2
(q) 80 2
(B) Radial acceleration, in
ms −2
(q) 5 5
2
(C) Radius of curvature, in m, at
t = 0 s.
(r) 25
(C) Acceleration, in ms −2
(r)
6
5
(D) Radius of curvature, in m, at
t = 1 s.
(s) 40
(D) Radius of curvature, in m
(s)
2
5
(E) Radius of curvature, in m, at
t = 2 s.
(t) 10
ms −1 , at t = 1 s.
(B) Magnitude of average
acceleration, in ms −2 , at t = 2 s.
(t) None of these
4.
2.
A particle is moving in a circle such that its speed
­varies with time t as v = ( 2t ) ms −1 . The quantities in
COLUMN-I are at t = 2 s against their values mentioned in COLUMN-II. Match them correctly.
Two inclined planes OA and OB having inclinations
30° and 60° with the horizontal respectively intersect each other at O, as shown in ­figure. A particle is
projected from point P with velocity u = 10 3 ms −1
along a direction perpendicular to plane OA . If the
particle strikes plane OB normally at Q . Based on the
information provided and taking g = 10 ms −2 , match
the quantities in COLUMN-I with the respective values in COLUMN-II.
COLUMN-I
COLUMN-II
(A) Distance travelled
(p) 2
(B) Displacement
(q) sin ( 2 )
(C) Average speed
(r) 4
(D) Average velocity
(s) 2 sin ( 2 )
A
(t) None of these
h
x
y
v
B
Q
u
P
60°
30°
O
05_Kinematics 2_Part 2.indd 73
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5.74 JEE Advanced Physics: Mechanics – I
COLUMN-I
COLUMN-II
(A) Time of flight, in s from
P to Q.
(p) 5
-1
(B) Velocity, in ms , with
which the particle strikes
the plane OB.
(q) 2
(C) Vertical height, in m, of
the point P above O
(r) 20
(D) Separation PQ, in m
(s) 10
(t) None of these
5.
6.
7.
A ball launched with some initial velocity from the origin moves in x -y plane such that x and y vary with
time t as x = α t and y = α t ( 1 − βt ) where α and
β are positive constants. Based on this information
match the quantities in COLUMN-I with the respective values in COLUMN-II.
COLUMN-I
COLUMN-II
(A) The maximum horizontal
distance travelled, as a multiple
α
of
2β
(p) 1
(B) The maximum vertical
displacement attained, as a
α
multiple of
16β
(q) 2
(C) The time taken by the ball to hit
the x-axis again, as a multiple
1
of
8β
(r) 4
(D) The acceleration of the ball, in
(s) 8
magnitude, as a multiple of
COLUMN-I
COLUMN-II
(A) Uniform motion
(p) Projectile motion
(B) Uniform accelerated
motion
(q) Uniform circular
motion
(C) Non uniform
accelerated motion
(r) Motion along a
straight line
(D) Uniform velocity
(s) Motion along ellipse
Two projectiles are launched with same initial speed
from ground at angles 30° and 60° . If R1 is range of
first and R2 is range of second, similarly H1 and H2
are their maximum heights and T1 and T2 are time
of flights, then match the ratios in COLUMN-I to the
values in COLUMN-II.
COLUMN-I
8.
COLUMN-II
(t) 16
1
3
(A)
R1
R2
(B)
H1
H2
(q) 1
(C)
T2
T1
(r)
3
(D)
T1H1R1
T2 H 2 R2
(s)
1
3 3
(p)
A body is projected with speed 20 2 ms −1 at an angle
45° with horizontal. After 1 s of it motion match the
following columns. g = 10 ms −2 .
(
αβ
8
(E) The velocity of the ball at half
the value of time calculated in
(C), as a multiple of a
Match the following
)
COLUMN-I
COLUMN-II
(A) Average velocity
(in magnitude)
(p) 10 5 ms −1
(B) Change in velocity
(in magnitude)
(q) 25 ms −1
(Continued)
05_Kinematics 2_Part 2.indd 74
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5.75
Chapter 5: Kinematics II
COLUMN-I
COLUMN-II
(C) Instantaneous speed
(r) 10 ms −1
10. Match the following
(D) Change in speed (nearly) (s) 6 ms −1
(in magnitude)
9.
Trajectory of particle launched obliquely from the
x2
, where, x and y are in
ground is given as y = x −
80
metre. For this projectile motion match the following if
g = 10 ms −2 .
COLUMN-I
COLUMN-II
(A) Angle of projection
(p) 20 m
(B) Angle of velocity with
horizontal after 4 s
(q) 80 m
(C) Maximum height
(r) 45°
(D) Horizontal range
1
(s) tan −1 ⎛⎜ ⎞⎟
⎝ 2⎠
COLUMN-I
COLUMN-II
(A) Particle moving in
circle
(p) a may be
perpendicular to v
(B) Particle moving in
straight line
(q) a may be in the
direction of v
(C) Particle undergoing
projectile motion
(r) a may make same
acute angle with v
(D) Particle moving into
space
(s) a may be opposite
velocity
11. A body is projected from the ground with velocity v at
an angle of projection θ . Then match the following.
COLUMN-I
COLUMN-II
(A) Change in momentum
(p) Remains
unchanged
(B) Angle at the highest
point
(q) Independent of
projected velocity
(C) Kinetic energy of body
(r) At highest point
is zero
(D) Horizontal component
of velocity
(s) Minimum at
highest point
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a numerical value obtained after doing series of calculations based on the data
given in the question(s).
1.
2.
A shell is fired from a gun from the bottom of a hill
along its slope. The slope of the hill is 30° and the
angle of the barrel to the horizontal 60° . The initial
velocity of the shell is 21 ms −1 . Find the distance, in
metre, from the gun to the point at which the shell
falls.
A particle is projected with velocity 2 gh so that it
just clears two walls of equal height h which are at
a distance 2h from each other. The time of passing
h
, where ∗ is not readable.
between the walls is ∗
g
Find ∗ .
05_Kinematics 2_Part 2.indd 75
3.
On a cricket field, the batsman is at the origin of co-ordinates and a fielder stands in position 46iˆ + 28 ˆj m .
(
)
The batsman hits the ball so that it rolls along the
ground with constant velocity 7.5iˆ + 10 ˆj ms −1 . The
iˆ
(
)
fielder can run with a speed of 5 ms −1 . If he starts to
run immediately the ball is hit what is the shortest
time, in seconds, in which he could intercept the ball?
4.
A particle is projected from a point at the foot of a fixed
plane, inclined at an angle of 45° to the horizontal, in
the vertical plane containing the line of greatest slope
through the point. If the particle strikes the plane
11/28/2019 7:19:53 PM
5.76 JEE Advanced Physics: Mechanics – I
­horizontally and ϕ ( > 45° ) is the angle of launch measured to the horizontal, then find the value of tan ϕ .
5.
6.
1
m with a
4
−1
linear speed of 2 ms . Calculate the angular speed of
a particle, in rads −1 .
A
u
3.9 m
B
A particle is moving in a circle of radius
C
6m
14.7 m
A boy whirls a stone in a horizontal circle of radius
0.5 m and at height 20 m above level ground. The
string breaks, and the stone flies off horizontally to
strike the ground after travelling a horizontal distance
of 10 m. Find the magnitude of the centripetal accel−2
eration, in ms , of the stone while in circular motion.
( Take g = 10 ms−2 )
7.
A highway curve is designed such that the cars travelling at a constant speed of 25 ms −1 must not have an
acceleration that exceeds 3 ms −2 . Determine the minimum radius of curvature, in metre, of the curve to the
nearest integer.
8.
At a given instant, a car travels along a circular curved
road with a speed of 20 ms −1 while decreasing its
speed at the rate of 3 ms −2 . If the magnitude of the
car’s acceleration is 5 ms −2 , determine the radius of
curvature of the road, in metre.
9.
1m
Ball bearings of diameter 20 mm leave the horizontal with a velocity of magnitude u and fall through
the 60 mm diameter hole at a depth of 800 mm as
shown. Calculate the permissible range of u , in cms −1
which will enable the ball bearings to enter the hole.
Take the dotted positions to represent the limiting conditions. Take g = 10 ms −2
(
20 mm
D
x
11. A particle is projected from a point at the foot of a fixed
plane, inclined at an angle of 45° to the horizontal, in
the vertical plane containing the line of greatest slope
through the point. If the particle strikes the plane at
right angles and ϕ ( > 45° ) is the angle of launch measured to the horizontal, then find the value of tan ϕ .
12. The minimum speed in ms −1 with which a projectile must be thrown from origin at ground so that
it is able to pass through a point P ( 30 m, 40 m ) is
( g = 10 ms−2 )
13. Two particles are simultaneously thrown from top of
two towers as shown. Their velocities are 2 ms −1 and
14 ms −1 . Horizontal and vertical separation between
these particles are 22 m and 9 m, respectively. Then
the minimum separation between the particles in process of their motion in meters is g = 10 ms −2 .
(
)
u = 2 ms–1
A
120 mm
45°
v = 14 ms–1
9m
45°
u
800 mm
60 mm
10. A ball is launched by a man standing on the top of a
building who holds the ball at a distance 1 m above
the edge A . Calculate the minimum velocity u ,
in ms −1 , along the horizontal such that the ball just
clears the edge C . Also find x , in metre, where the
ball strikes the ground. Take g = 9.8 ms −2 .
05_Kinematics 2_Part 2.indd 76
)
B
22 m
14. A projectile is fixed at an angle 60° with horizontal.
Ratio of initial and final kinetic energy velocity vector
of projectile makes an angle 15° with velocity of projection is
15. A particle is projected from the bottom of an inclined
plane of inclination 30° . At what angle α (from the
horizontal), in degree, should the particle be projected
to get the maximum range on the inclined plane?
11/28/2019 7:19:59 PM
Chapter 5: Kinematics II
16. A small body is released from point A of smooth parabolic path y = x 2 , where y is vertical axis and x is
horizontal axis at ground as shown. The body leaves
the surface from point B . If g = 10 ms −2 , then total
horizontal distance in meters travelled by body before
it hits ground is _______
Y
α
O
u=0
–2 m
x
1m
17. A particle is projected towards north with speed
20 ms −1 at an angle 45° with horizontal. Ball get
horizontal acceleration of 7.5 ms −2 towards east due
to wind. Range of ball (in meter) minus 42 m will be
18. For an observer on trolley, direction of projection
of particle is shown in figure, while for observer on
ground ball rises vertically. Maximum height (in
meter) reached by ball minus 10 m is ______
v
(w.r.t. trolley)
60°
A
10 m
3
37°
X
21. A particle is projected from ground with minimum
speed required to hit a target at a height h = 10 m at
B
0
B
C
u
y
A
5.77
10 ms–1
a horizontal distance d = 300 m as shown. Then find
the time taken by particle (in seconds) to hit the target.
g = 10 ms −2
(
Target
h
d
22. A particle is projected with initial velocity
v = 10 2 ms −1 as shown. After elastic collision with
the inclined plane, the particle rebounds normally
with the plane and retraces its path to come back at
its point of projection. Then find the time in seconds
in which particle returns to the point of projection.
g = 10 ms −2
(
19. Two seconds after projection, a projectile is travelling
in a direction inclined at 30° with horizontal. After
one more second, it is travelling ­
horizontally. One
tenth of the angle of projection (in degree) with horizontal is _____
20. A particle is projected from O on the ground with
⎛ 1⎞
velocity u = 5 5 ms −1 at angle α = tan −1 ⎜ ⎟ .
⎝ 2⎠
It strikes at a point C on a fixed plane AB having
inclination of 37° with horizontal as shown, then the
x-coordinate of point C in meters is g = 10 ms −2
(
)
)
v
β = sin–1
1
3
23. A football is thrown with a velocity of 10 ms −1 at an
angle of 30° above the horizontal. What will the time
of flight, in seconds? g = 10 ms −2
(
)
)
ARCHIVE: JEE MAIN
1. [Online April 2019]
A plane is inclined at an angle 30° with respect to the
horizontal. A particle is projected with a speed 2 ms −1,
from the base of the plane, making an angle 15° with
respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane
is close to (Take g = 10 ms −2 )
05_Kinematics 2_Part 2.indd 77
u
15°
30°
(A) 18 cm
(B)
(C) 14 cm
(D) 26 cm
20 cm
11/28/2019 7:20:06 PM
5.78 JEE Advanced Physics: Mechanics – I
2. [Online April 2019]
A shell is fired from a fixed artillery gun with an initial
speed u such that it hits the target on the ground at a
distance R from it. If t1 and t2 are the values of the
time taken by it to hit the target in two possible ways,
the product t1t2 is
(A)
R
2g
(B)
2R
g
(C)
R
g
(D)
R
4g
(
−2
)
2 ⎞
3
(B) θ 0 = cos ⎜
and v0 = ms −1
⎝ 5 ⎟⎠
5
−1 ⎛
5
⎛ 1 ⎞
and v0 = ms −1
(C) θ 0 = cos −1 ⎜
⎝ 5 ⎟⎠
3
3
⎛ 2 ⎞
and v0 = ms −1
(D) θ 0 = sin −1 ⎜
⎝ 5 ⎟⎠
5
4. [Online April 2019]
Two particles are projected from the same point with
the same speed u such that they have the same range
R, but different maximum heights, h1 and h2 . Which
of the following is correct?
(A) R2 = 4 h1h2
(B)
(C) R2 = 2 h1h2
(D) R2 = h1h2
R2 = 16 h1h2
5. [Online January 2019]
Two guns A and B can fire bullets at speeds 1 kms −1
and 2 kms −1 respectively. From a point on a horizontal ground, they are fired in all possible directions. The
ratio of maximum areas covered by the bullets fired by
the two guns, on the ground is
(B) 1 : 8
(D) 1 : 16
6. [Online 2015]
If a body moving in a circular path maintains constant speed of 10 ms −1 , then which of the following
correctly describes relation between acceleration and
radius?
(D) a
r
7. [2013]
A projectile is given an initial velocity of iˆ + 2 ˆj ms −1,
where iˆ is along the ground and ˆj is along the vertical. If g = 10 ms −2 , the equation of its trajectory is
iˆ
iˆ
(
)
iˆ
(A) 4 y = 2x − 25x 2
(B)
(C) y = 2x − 5x 2
(D) 4 y = 2x − 5x 2
y = x − 5x 2
8. [2012]
A boy can throw a stone up to a maximum height of
10 m . The maximum horizontal distance that the boy
can throw the same stone up to will be
(A) 10 m
(B) 10 2 m
(C) 20 m
(D) 20 2 m
9. [2011]
A water fountain on the ground sprinkles water all
around it. If the speed of water coming out of the fountain is v , the total area around the fountain that gets
wet is
v2
g
(B) π
v4
g2
π v4
2 g2
(D) π
v2
g2
(A) π
(C)
10. [2010]
A small particle of mass m is projected at an angle θ
with the x-axis with an initial velocity v0 in the x -y
plane as shown in the figure. At a time t <
angular momentum of the particle is
v0 sin θ
, the
g
y
v0
θ
05_Kinematics 2_Part 2.indd 78
r
r
5
⎛ 1 ⎞
(A) θ 0 = sin −1 ⎜
and v0 = ms −1
⎝ 5 ⎟⎠
3
(A) 1 : 4
(C) 1 : 2
r
(C) a
3. [Online April 2019]
The trajectory of a projectile near the surface of the
earth is given as y = 2x − 9x 2 . If it were launched at an
angle θ 0 with speed v0 then g = 10 ms
(B) a
(A) a
x
11/28/2019 7:20:14 PM
Chapter 5: Kinematics II
1
mgv0t 2 cos θ iˆ
2
(C) mgv0t cos θ kˆ
(A)
(B) − mgv0t 2 cos θ ˆj
iˆ
iˆ
1
(D) − mgv0t 2 cos θ kˆ
2
iˆ
iˆ
where iˆ , ˆj and k̂ are unit vectors along x , y and
z-axis respectively.
iˆ
iˆ
12. [2010]
A point P moves in counter-clockwise direction on a
circular path as shown in the figure. The movement of
P is such that it sweeps out a length s = t 3 + 5 , where
s is in metres and t is in seconds. The radius of the
path is 20 m . The acceleration of P when t = 2 s is
nearly
y
11. [2010]
For a particle in uniform circular motion, the accelera
tion a at a point P ( R, θ ) on the circle of radius R is
(Here q is measured from the x-axis)
v2 ˆ v2 ˆ
i+
j
(A)
R
R
(C) −
iˆ
B
P(x, y)
20
v2
v2
(B) − cos θ iˆ + sin θ ˆj
R
R
iˆ
O
iˆ
v2
v2
sin θ iˆ + cos θ ˆj
R
R
(D) −
iˆ
v2
v2
cos θ iˆ − sin θ ˆj
R
R
5.79
m
A
x
(A) 14 ms −2
(B) 13 ms −2
(C) 12 ms −2
(D) 7.2 ms −2
ARCHIVE: JEE advanced
Integer/Numerical Answer Type Questions
In this section, the answer to each question is a
numerical value obtained after series of calculations
based on the data provided in the question(s).
1. [JEE (Advanced) 2019]
A ball is thrown from ground at an angle θ with horizontal and with an initial speed u0 . For the resulting
projectile motion, the magnitude of average velocity of
the ball up to the point when it hits the ground for the first
time is V1 . After hitting the ground, the ball rebounds
u
at the same angle θ but with a reduced speed of 0 .
α
Its motion continues for a long time as shown in Figure.
u0
θ
u0 /α
u0 /α 2
θ
θ
u0 /α m
θ
If the magnitude of average velocity of the ball for
entire duration of motion is 0.8 V1, the value of a is
……..
05_Kinematics 2_Part 2.indd 79
2. [JEE (Advanced) 2018]
A ball is projected from the ground at an angle of 45°
with the horizontal surface. It reaches a maximum
height of 120 m and returns to the ground. Upon
hitting the ground for the first time, it loses half of
its kinetic energy. Immediately after the bounce, the
velocity of the ball makes an angle of 30° with the
horizontal surface. The maximum height it reaches
after the bounce, in metres, is ______.
3. [JEE (Advanced) 2011]
A train is moving along a straight line with a constant
acceleration a . A boy standing in the train throws a
ball forward with a speed of 10 ms −1 , at an angle of
60° to the horizontal. The boy has to move forward
by 1.15 m inside the train to catch the ball back at the
initial height. The acceleration of the train, in ms −2 , is
11/28/2019 7:20:19 PM
5.80 JEE Advanced Physics: Mechanics – I
Answer Keys—Test Your Concepts and Practice Exercises
Test Your Concepts-I
(Based on Curvilinear Motion)
7. 3.16 ms −1
3. 8 rads −1
9. (a) 5 s
(b) 130 m
4. 2 rads −2
(c) 43.6 ms −1
5. 9.5 ms −2
12. t =
6. 50 ms −2
uv sin ( α − β )
v
( cos β − u cos α )
7. 20 s, 0.32 ms −2
13. 4.5 m, 2.75 m
8. 1.8 ms −1 , 1.2 ms −2
15. 68.2° , 1.253 ms −1
9. 38.7 ms
−1
16. 28 ms −1
−1
10. 63 ms
17.
11. 0.488 ms −2
12. 208 m
13. 100 m
14. 3.2 m
gt cos β
2 sin ( α − β )
21. tan α =
Test Your Concepts-II
(Based on Horizontal Projectile)
g ( R1 − R2 )
2
4v 2 ( R1 + R2 )
Test Your Concepts-IV
(Based on Projectile on an Inclined Plane)
1. 44.1 m, 29.4 ms −1
1. (a) 2
(b) 3
−1
2. 20 ms
3. 0.7 s, 7 m
3.
2u2 tan θ sec θ
g
4.
2 gh
2 + cot 2 θ
7. uMIN = 25 cms −1 and uMAX = 35 cms −1
5.
Test Your Concepts-III
(Based on Oblique Projectile)
gR ( 1 + 3 sin 2 β )
2 sin β
6. α + β =
π
2
4. 14 ms
−1
5. 34 ms
, 2.4 m
−1
6. 6 ms −1 , ( 1 + 3.9 + 14.7 ) =
1
( 9.8 ) t 2 , 6 m
2
1. 76°
6. (a) 11 m
(b) 23 m
(c) 17 ms −1 , 63° with horizontal
Single Correct Choice Type Questions
1. D
2. D
3. D
4. D
5. D
6. B
7. A
8. C
9. A
10. D
11. A
12. A
13. C
14. B
15. C
16. D
17. B
18. B
19. A
20. D
21. C
22. D
23. D
24. D
25. D
26. C
27. A
28. A
29. D
30. B
31. C
32. D
33. B
34. A
35. D
36. D
37. B
38. A
39. D
40. A
05_Kinematics 2_Part 2.indd 80
11/28/2019 7:20:23 PM
5.81
Chapter 5: Kinematics II
41. B
42. C
43. C
44. D
45. C
46. C
47. B
48. A
49. C
50. C
51. D
52. D
53. C
54. D
55. B
56. C
57. B
58. D
59. B
60. B
61. A
62. D
63. C
64. C
65. D
66. C
67. B
68. C
69. D
70. D
71. B
72. B
73. C
74. A
75. D
76. C
77. D
78. A
79. C
80. C
81. A
82. D
83. D
84. B
85. C
86. C
87. B
88. C
89. A
90. B
91. C
92. B
93. B
94. C
95. A
96. C
97. B
98. D
99. D
100. C
101. C
102. A
103. D
104. B
105. D
106. C
107. B
108. C
109. C
110. D
111. B
112. B
113. A
114. A
115. D
116. C
117. B
118. B
119. B
120. D
121. C
122. C
123. B
124. C
125. D
126. D
Multiple Correct Choice Type Questions
1. B, C, D
2. A, C
3. C, D
4. A, D
5. A, D
6. C, D
7. A, B, C, D
8. A, B, C
9. A, D
10. B, D
11. A, B, D
12. A, B, C, D
13. B, C
14. C, D
15. A, B, C, D
16. B, D
17. A, B, C, D
18. A, B, D
19. B, C
20. A, C, D
21. A, B, C, D
22. A, D
23. B, C
24. B, C
25. A, B
26. A, B
27. B, C
Reasoning Based Questions
1. B
2. D
3. D
11. C
12. A
13. A
4. D
5. D
6. A
7. B
8. D
9. A
10. A
Linked Comprehension Type Questions
1. C
2. B
3. D
4. A
5. D
6. C
7. B
8. B
9. D
10. A
11. D
12. B
13. B
14. B
15. C
16. C
17. C
18. C
19. D
20. D
21. D
22. D
23. D
24. B
25. D
26. B
Matrix Match/Column Match Type Questions
1. A → (r)
B → (t)
C → (q)
D → (p)
2. A → (r)
B → (s)
C → (p)
D → (q)
3. A → (r)
B → (s)
C → (p)
D → (q)
4. A → (q)
B → (s)
C → (p)
D → (r)
5. A → (q)
B → (r)
C → (s)
D → (t)
6. A → (q, r)
B → (p, r)
C → (q, r, s)
D → (r)
7. A → (q)
B → (p)
C → (r)
D → (s)
8. A → (q)
B → (r)
C → (p)
D → (s)
9. A → (r)
B → (r)
C → (p)
D → (q)
10. A → (p, r)
B → (q, s)
C → (p, q, r, s)
D → (p, q, r, s)
11. A → (q)
B → (r)
C → (s)
D → (p)
05_Kinematics 2_Part 2.indd 81
E → (s)
E → (p)
11/28/2019 7:20:23 PM
5.82 JEE Advanced Physics: Mechanics – I
Integer/Numerical Answer Type Questions
1. 30
2. 2
3. 4
4. 2
5. 8
6. 50
7. 208
8. 100
9. Min. 25 & Max. 35 10. 6, 6
11. 3
12. 30
13. 6
14. 2
15. 60
16. 8
17. 8
18. 5
19. 6
20. 5
21. 2
22. 6
23. 1
ARCHIVE: JEE MAIN
1. B
2. B
11. D
12. A
3. C
4. B
5. D
6. C
7. C
8. C
9. B
10. D
ARCHIVE: JEE Advanced
Integer/Numerical Answer Type Questions
1. 4.00
05_Kinematics 2_Part 2.indd 82
2. 30
3. 5
11/28/2019 7:20:23 PM
CHAPTER
6
Newton’s Laws of
Motion
Learning
Objectives
After reading this
chapter, you will be able to:
After reading this chapter, you will be able to understand concepts and problems based on:
(a) Newton’s Laws of Motion
(c) Friction
(b) Pseudo Force
(d) Dynamics of Circular Motion
All this is followed by a variety of Exercise Sets (fully solved) which contain questions as per the
latest JEE pattern. At the end of Exercise Sets, a collection of problems asked previously in JEE (Main
and Advanced) are also given.
dYNAMIcS: AN INTrOducTION
In kinematics, we studied the motion of a particle,
with emphasis on motion along a straight line and
motion in a plane and simply described it in terms
of vectors r , v and a. Now comes the time when
we shall be discussing about the cause producing
motion. This treatment, happens to be an aspect of
mechanics, known as dynamics. In this chapter, we
shall be primarily discussing the forces along with
their respective nature and properties that account for
the motion of a body. As done before, the bodies will
be treated as if they were single particles. However, in
the later chapters, we shall be extending the properties of this chapter to discuss the motion of a group of
particles and extended bodies as well. So, this branch
of Physics dealing with motion along with the cause
producing motion is called Dynamics.
FOrcE
Force is a pull or push which changes or tends
to change the state of rest or of uniform motion or
06_Newtons Laws of Motion_Part 1.indd 1
direction of motion of any object. Force is the interaction between the object and the source (providing
the pull or push). It is a vector quantity. Its SI unit is
newton (N) and cgs unit is dyne
1 N = 10 5 dyne
A force acting on a body
(a) may change only speed.
(b) may change only direction of motion.
(c) may change both the speed and direction of
motion.
(d) may change size and shape of a body.
Unit of Force
SI unit of force is newton (N) and 1 N = 1 kgms -2
cgs unit of force is dyne (dyn) and 1 dyne = 1 gcms -2
Also, 1 newton = 10 5 dyne .
The dimensional formula of force is [ MLT -2 ]
another commonly used unit of force is kilogram force (kgf). It is the force with which a body
of mass 1 kg is attracted towards the centre of the
11/28/2019 7:26:39 PM
6.2 JEE Advanced Physics: Mechanics – I
earth. So 1 kgf can also be thought as the weight
­corresponding to a body of mass 1 kg is the weight
1 kgf = 9.8 N and 1 gf = 980 dyne
Conceptual Note(s)
A force is a vector quantity. Since we know that a vector
quantity is a quantity which has both magnitude and
direction. To fully describe the force acting upon an
object, you must describe both the magnitude (size or
numerical value) and the direction. Thus, 10 N, is not a
full description of the force ­acting upon an object. In
contrast, 10 N, downwards is a complete description
of the force acting upon an object because, both the
magnitude (10 N) and the direction (downwards) are
given.
The behaviour of all objects can be described by
saying that objects tend to keep on doing what they
are doing (unless acted upon by an unbalanced force).
There are two parts of the statement to Newton’s
First Law.
(a) one which predicts the behaviour of stationary
objects and
(b) the other which predicts the behaviour of
moving objects.
These two parts are summarized in the following
­diagram.
Forces are Balanced
Objects at Rest
(v = 0)
Objects in Motion
(v ≠ 0)
a=0
a=0
Newton gave three laws of motion, based on which
motion associated with a particle can be explained
easily.
Stay at Rest
Stay in Motion
(Same speed and direction)
(a) The First Law or The Law of Inertia
(b) The Second Law ( F = ma )
(c) The Third Law or The Action - Reaction Law
Conceptual Note(s)
NEWTON’S LAWS OF MOTION
Newton’s First Law or Law of Inertia
A body continues to maintain its state of rest or of
uniform motion or of direction unless and until
some external unbalanced force acts on it. If there is
an interaction between the body and objects present
in the environment, the effect may be to change the
“natural” state of the body’s motion.
Examples:
(a) When a bus suddenly starts/stops, the passengers
tends to move backward/forward.
(b) A carpet is beaten with a stick to remove its dust.
(c) A coin kept on a cardboard on a glass tumbler, on
striking hard the cardboard, the coin falls into the
glass tumbler.
(d) Athlete runs some distance before making a jump.
In other words, an object at rest tend to stay at rest
and an object in motion tend to stay in motion with
the same speed and in the same direction unless acted
upon by an unbalanced force.
06_Newtons Laws of Motion_Part 1.indd 2
It has been a common observation that every object
resists any effort to change its velocity (both in
magnitude and direction). This property expressing
the degree of insusceptibility of a body to any change
in its velocity is called the property of inertia.
Different bodies reveal this property in different
degrees. A measure of inertia is provided by the
quantity called mass. A body possessing a greater
mass has more inertia i.e., if equal forces are applied
on two different bodies, the body with the greater
mass possesses smaller acceleration.
Momentum ( p )
It is observed that more the mass of the body, the
more is the momentum possessed by it. Similarly,
the more the velocity possessed by a body, the more
is its momentum. In simpler words, actually the
momentum is a measure of the amount of motion
possessed by a body. It is defined as the product of
mass and velocity or momentum is just equal to
the mass times velocity. The momentum is always
11/28/2019 7:26:40 PM
Chapter 6: Newton’s Laws of Motion
directed along the velocity of the body whose
momentum is being measured. Mathematically,
p = mv
It is a vector quantity with SI unit kgms -1 and
dimensional formula MLT -1.
Newton’s Second Law of Motion
The external force applied on a body is equal to the
rate of change of momentum of a body.
dp
Fext = F =
dt
dp d
F=
= ( mv )
dt dt
For constant mass system(s)
dv
F=m
= ma
dt
For variable mass system(s)
dm
dv
+m
F=v
dt
dt
Special Case
dv If v = constant , then
=0
dt
dm
⇒ F =v
dt
6.3
Conceptual Note(s)
(a) The Second Law is obviously consistent with the
First Law as F = 0 implies a = 0.
(b) The Second Law of motion is a vector law. It is
actually a combination of three equations, one
for each component of the vectors.
dp x
= ma x
dt
dp y
⇒ Fy =
= ma y
dt
dpz
⇒ Fz =
= ma z
dt
This means that if a force is not parallel to the
velocity of the body, but makes some angle with
it, it changes only the component of velocity
along the direction of force. The component of
velocity normal to the force remains unchanged.
(c) The Second Law of motion given above is strictly
applicable to a single point mass. The force F in
the law stand for the net external force on the
particle and a stands for the acceleration of the
particle. Any internal forces in the system are not
to be included in F.
Fx =
Newton’s Third Law of Motion
Solution
To every action there is equal and opposite reaction
and both must act on two different bodies. In other
words, all interaction forces exist in pair, a pair which
has two forces of equal magnitude, opposite direction
and acting on different bodies along a straight line
connecting them. Also the two forces comprising the
pair have the same nature.
i.e., FAB = - FBA …(1)
FAB = Force on body A due to B
FBA = Force on body B due to A
From Second Law of Motion
dp d
F=
= ( mv )
dt dt
⇒
mA aA = mB aB ⇒
aB =
dm
Thrust = F = v
dt
e.g., Rocket propulsion
Illustration 1
The velocity of a particle of mass 2 kg is given by
v = atiˆ + bt 2 ˆj . Find the force acting on the particle.
⇒
⇒
(
)
d
atiˆ + bt 2 ˆj
dt
ˆ
ˆ
F = 2 ai + 4btj F=2
06_Newtons Laws of Motion_Part 1.indd 3
From (1), we get
FAB = FBA
mA a A
mB
If mA mB , then aB → 0 (Think of apple-earth
example)
11/28/2019 7:26:45 PM
6.4 JEE Advanced Physics: Mechanics – I
Example:
A block is being pulled by means of a rope. I have characterised some a­ ction-reaction pairs and have shown them
in figures. Please see these figures carefully.
FRB
FBR
FRM
FMR
Block
Man
FBR
FRB
Rope
FRM
FMR
We also have, FBR = -FRB and FRM = - FMR
Conceptual Note(s)
The negative sign tells about the attractive nature of
the gravitational force as the test mass is always attracted
towards the source mass (i.e. opposite to r ).
(i) It is the weakest force and is always attractive.
(ii) It is a long range force as it acts between any two
particles situated at any distance in the universe.
(iii) Gravitational force between two bodies is
independent of the presence of other bodies and
it is also independent of the nature of intervening
medium (i.e. medium present between the
bodies).
(iv) It is a central force i.e. it acts along the line joining
the centres of the two bodies.
(v) It is negligible for lighter bodies, however it
dominates in case of planetary bodies and their
motion.
(vi) Gravitons are exchange particles between two
bodies and are responsible for the gravitational
interaction between them.
Action reaction forces always act on different bodies
and their line of action is the same.
FUNDAMENTAL FORCES IN NATURE
All the forces observed in nature such as muscular
force, tension, reaction, friction, elastic, weight, electric, magnetic, nuclear etc., can be explained in terms
of only following four basic interactions.
Gravitational Force and Properties
The force of interaction which exists between two
particles due to their masses is called gravitational
force. The gravitational force between two masses
m1 (called the source mass S) and m2 (called the test
mass T) separated by a distance r is given by
S
r
Earth
Consider an apple to be falling freely as shown in
­figure. When it is at a height h, force between earth
and apple is given by
F=
GmMe
r
2
=
GmMe
( Re + h )2
where Me is the mass of earth, Re is the radius of earth.
It acts towards earth’s centre. Now rearranging above
result, we get
T
mm F = -G 1 3 2 r
r
where, r is position vector of test particle T with
respect to source particle S and G is universal
gravitational constant whose value is given by
G = 6.67 × 10 -11 Nm 2 kg -2 .
06_Newtons Laws of Motion_Part 1.indd 4
Apple
h
⎛ GM ⎞ ⎛ Re ⎞
F = m⎜ 2 e ⎟ ⎜
⎟
⎝ Re ⎠ ⎝ Re + h ⎠
2
2
⇒
⎛ Re ⎞
F = mg ⎜
⎝ Re + h ⎟⎠
For h Re , we have
GMe ⎫
⎧
⎨∵ g = 2 ⎬
Re ⎭
⎩
Re
1
Re + h
11/28/2019 7:26:48 PM
Chapter 6: Newton’s Laws of Motion
⇒
F = mg This is the force exerted by earth on any particle of
mass m near the earth surface.
The value of g = 9.81 ms-2  10 ms-2  p2 ms-2
 32 fts-2. It is also called acceleration due to gravity
near the surface of earth.
Electromagnetic Force
It includes the electrical and the magnetic forces
which may be attractive or repulsive. Photons are
exchange particles for electromagnetic interactions.
So, force exerted by one particle on the other because
of the electric charge on the particles is called electromagnetic force.
Following are the main characteristics of electromagnetic force
(a) These can be attractive or repulsive.
(b) These are long range forces.
(c) These depend on the nature of medium between
the charged particles.
(d) All macroscopic forces (except gravitational)
which we experience as push or pull or by contact
are electromagnetic, i.e., tension in a rope, the
force of friction, normal reaction, muscular force,
and force experienced by a deformed spring are
electromagnetic forces. These are manifestations
of the electromagnetic attractions and repulsions
between the atoms or molecules.
Nuclear Force
An attractive force, it is the force between two nucleons. It is the strongest force that binds all the nucleons
in a tiny volume (corresponding to radius of order
of 1 fermi) inspite of large electric repulsion between
protons. Mesons are the exchange particles responsible for nuclear interactions. It acts within the nucleus
that too upto a very small distance. Radioactivity,
f­ission and fusion etc. result because of unbalancing
of nuclear forces.
negative beta decay by ­emitting an electron and a
particle called antineutrino. It is not the same force
as gravitational, electromagnetic or nuclear force.
The range of weak force is very small, in fact much
smaller than the size of a proton or a neutron.
It has been found that for two protons at a distance
of 1 fermi
FN : FEM : FW : FG :: 1 : 10 -2 : 10 -7 : 10 -38
CLASSIFICATION OF FORCES ON THE BASIS
OF CONTACT
On the basis of contact, forces can be classified as
Field Force
Force which acts on an object at a distance by the
interaction of the object with the field produced
by other object is called Field Force or Action at a
Distance Force.
Examples:
(a) Gravitation force
(b) Electromagnetic force
Contact Force
Forces which are transmitted between bodies by
short range atomic molecular interactions are called
contact forces. When two objects come in contact they
exert contact forces on each other.
Examples:
Normal Force, Mechanical Force, Force of Friction
Normal force (N) or Normal Reaction (N)
It is the component of contact force perpendicular to the
surface. It measures how strongly the surfaces in contact
are pressed against each other. It is the electromagnetic
force.
A table is placed on Earth as shown in figure.
Weak Force
It exists between elementary particles and is
responsible for the change in nucleus . Under its
action a neutron can change into proton during
06_Newtons Laws of Motion_Part 1.indd 5
6.5
2
1
3
4
Here table presses the earth so normal force exerted by
four legs of table on earth are as shown in figure.
11/28/2019 7:26:49 PM
6.6 JEE Advanced Physics: Mechanics – I
N1
N3
Real Force
N2
N4 ground
Now a boy pushes a block kept on a frictionless ­surface.
Here, force exerted by boy on block is electromagnetic
interaction which arises due to similar charges appearing on finger and contact surface of block, it is normal
force.
Block
(by boy) N
Block
A block is kept on inclined surface. Component of its
weight presses the surface perpendicularly due to which
contact force acts between surface and block. Normal
force exerted by block on the surface of inclined plane
is shown in figure. Force acts perpendicular to the
surface.
N
Force which acts on an object due to other object is
called as real force.
An isolated object (far away from all objects) does not
experience any real force.
CONCEPT OF IMPULSE
AND IMPULSE AS
AREA UNDER F-t GRAPH
Whenever a large force ( F ) acts on a body for an
extremely small time (say dt), then we introduce the
concept of Impulse ( I ). So, impulse is just defined as
the product of the large force with the small time.
⇒ dI = Fdt Impulse is also defined as the integral of force with
respect to time.
I=
tf
∫ Fdt
ti
θ
θ
Frictional Force (f)
A contact force, it arises due to a contact between the
surfaces of the two bodies and comes into being when
there is a relative motion between the surfaces in contact. Actually, it is the component of the reaction force
tangential to the surface on which the body is kept.
Since force is a vector and time is a scalar, the result of
the integral in above equation is a vector. If the force
is constant (both in magnitude and direction), it may
be removed from the integral so that the integral is
reduced to
tf
∫
I = F dt = F ( t f - ti ) = F Δt
ti
SYSTEM
Two or more than two objects which interact with
each other form a system. The classification of forces
on the basis of boundary of system are given.
Graphically, the impulse is the area between the force
curve and the F = 0 axis, as shown in figure.
F
F
d
c
a
b
Internal Forces
Forces acting each with in a system among its
constituents.
External Forces
Forces exerted on the constituents of a system by the
outside surroundings are called as external forces.
06_Newtons Laws of Motion_Part 1.indd 6
t
dt
The SI unit of impulse is Ns.
If more than one force are acting on a particle,
then the net impulse is given by the time integral of
the net force.
11/28/2019 7:26:51 PM
Chapter 6: Newton’s Laws of Motion
tf
I net =
∫
Solution
Fnet dt
ti
For example: A cricketer hitting a ball to score a six.
The bat hits the ball with the large force for a very
short
time.
If F be the large force acting on a body for a small
time dt, then
dI = Fdt
⇒
I=
∫
dI =
t2
∫
∫
Illustration 2
Find the impulse due to the force F = a iˆ + bt ˆj , where
a = 2 N and b = 4 Ns -1, if this force acts from ti = 0 to
t f = 0.3 s.
Solution
tf
∫
Fdt =
⇒
⇒
∫
∫ tdt
0
0.3
bt 2 ˆ
I = at iˆ
+
j
0
2 0
An 150 g ball is thrown at 30 ms -1 . It is struck by a
bat, which gives it a velocity of 40 ms -1 in the opposite direction. If the time of contact is 10 -2 s, what is
the average force on the ball?
Solution
If we choose the original direction as +x-axis, then
Δp = mv f - mvi = m ( -40iˆ - 30iˆ )
Fav =
IMPULSE – MOMENTUM THEOREM
According to Newton’s Second Law, we have
dp
F=
dt
⇒ dp = Fdt ( 4 ) ( 0.3 )2 ˆ
I = ( 2 ) ( 0.3 ) iˆ +
j
2
ˆ
ˆ
I = 0.6i + 0.18 j Ns
(
)
f
⇒
Illustration 3
(
)
A ball falling with velocity vi = -0.65 iˆ - 0.35 ˆj ms -1
is subjected to a net impulse I = 0.6iˆ + 0.18 ˆj Ns -1.
iˆ
iˆ
(
)
If the ball has a mass of 275 g, calculate its velocity
immediately following the impulse.
06_Newtons Laws of Motion_Part 1.indd 7
Δp -0.15 kg × 70 iˆ
=
= -1050 iˆ N
Δt
10 -2
Notice that this is much larger than the weight (1.5 N)
of the ball.
iˆ
iˆ
iˆ
iˆ
0.3
dt + b ˆj
)
Illustration 4
∫ ( a iˆ + bt ˆj )dt
0.3
⇒
(
) (
v f = ( 1.53iˆ + 0.305 ˆj ) ms -1
f
iˆ
The average force is
0.3
0.3
0
⇒
iˆ
0
ti
I = a iˆ
Using Impulse - Momentum Theorem
mv f - mvi = I
I
⇒ v f = vi +
m
0.6 iˆ + 0.18 ˆj
Thus, v f = -0.65iˆ - 0.35 ˆj +
0.275
⇒
v = -0.65iˆ - 0.35 ˆj + 2.18iˆ + 0.655 ˆj
⇒
F dt
t1
Since dI = Area abcd = Fdt
I = Fdt = Area under a curve in a F -t graph.
⇒
I=
6.7
∫
⇒
f
∫
Fdt = I
i
I = pfinal - pinitial
i
⇒
dp =
Impulse = Change in momentum
{called Impulse Momentum Theorem}
⇒ I = m(v - u ) 11/28/2019 7:27:00 PM
6.8 JEE Advanced Physics: Mechanics – I
Conceptual Note(s)
In the following cases shown, the change in
momentum is calculated.
Change in Momentum
Δp = p f - pi
y
x
cASE-1
m
Initially
u
along +x direction
m
cASE-2
m
Initially
u
m
Finally
u
cASE-3
θ
m
Initially
v
θ
m
Finally
A box is put on a scale which is adjusted to read zero,
when the box is empty. A stream of pebbles is then
poured into the box from a height h above its bottom
at the rate of n pebbles per second. Each pebble has a
mass m. If the pebbles collide with the box such that
they immediately come to rest after the collision. Find
the reading at time t after which the pebbles begin to
fill the box.
SOLuTION
Velocity (v) of each pebble just before striking the box
is v = 2gh
v
Finally
v
Δp = m ( v - u ) ,
ILLuSTrATION 5
Δp = -m ( v + u )
⇒ Δp = m ( v + u ),
along -x direction
Δp x = m ( v x - ux )
⇒ Δp x = -m ( v cos ϕ + u cosq ) ,
along -x direction
( Δp )y = m ( v y - uy )
⇒ ( Δp ) y = m ( v sinϕ - u sinq ) ,
along +y direction.
So, Δp = ( Δp x ) iˆ + ( Δp y ) ˆj
⎛ The additional ⎞
⎛ Sum of ⎞ ⎜
force exerted ⎟
⎛ Reading ⎞
⎜ weights of ⎟ ⎜
⎜ of the ⎟ ⎜
by
the pebbles ⎟
⎟
⎟
⎜ scale at ⎟ = ⎜ pebbles ⎟ + ⎜
due to a
⎜
⎟
⎜
⎟
collected ⎟ ⎜
change in ⎟
⎝ time t ⎠ ⎜⎜
⎟
⎝ in time t ⎠ ⎜
⎟
⎝ momentum ⎠
Total weight of pebbles in time t is (mg)(nt)
Change in momentum of 1 pebble is m
( 2 gh )
Change in momentum (Δp) of nt pebbles is
(
)
Δp = m 2 gh ( nt )
⇒ Additional force =
(
Δp
= mn 2 gh
t
⇒ Total force = mn gt + 2 gh
)
Test Your Concepts-I
Based on Impulse Momentum
(Solutions on page H.193)
30
°
y
x
30
°
1. A steel ball of mass 3 kg strikes a wall with a speed
of 10 ms -1 at an angle of 30° with the surface. It
bounces off with the same speed and angle as
shown. If the ball is in contact with the wall for
200 ms, calculate the average force exerted by the
wall on the ball.
06_Newtons Laws of Motion_Part 1.indd 8
11/28/2019 7:27:06 PM
Chapter 6: Newton’s Laws of Motion
2. A ball of mass 150 g is dropped from a height of
20 m. It rebounds from the floor to reach a height
of 5 m. What impulse was given to the ball by the
floor? Take g = 10 ms -2 .
3. An estimated force-time curve for a ball struck by a
bat is shown in figure. From the curve, determine.
(a) the impulse delivered to the ball
(b) the average force exerted on the ball and
(c) the peak force exerted on the ball.
F(N)
F = 16000 N
20000
6.9
4. A garden hose is originally full of motionless water.
What additional force is necessary to hold the
nozzle stationary after the water flow is turned on,
if the discharge rate is 0.6 kgs -1 with a speed of
25 ms -1 ?
5. A professional driver of mass 60 kg performs a dive
from a platform 10 m above the water surface.
Find the magnitude of the average impact force
experienced by him if the impact time is 1 s on
collision with the water surface. Assume that the
velocity of the diver just after ­entering the water
surface is 4 ms -1. Take g = 9.8 ms -2 .
15000
10000
5000
0
1
2
3
4
t(ms)
CONSTRAINED MOTION OF CONNECTED
PARTICLES
Sometimes it is observed that the motions of particles
are interrelated because of the constraints imposed by
interconnecting members. In such cases it becomes necessary for us to account for these constraints in order to
determine the respective motions of the particles.
One Degree of Freedom
Let us consider a simple system of two interconnected particles A and B shown in figure. It is quite
evident by inspection that the horizontal motion of
A is twice the vertical motion of B. However, we will
use this example to illustrate the method of analysis
which applies to more complex situations where the
results cannot be easily obtained by inspection.
DATUM 1
x
A
r2
b DATUM 2
y
r1
B
06_Newtons Laws of Motion_Part 1.indd 9
The motion of B is clearly the same as that of the
center of its pulley, so we establish position coordinates y and x measured from a convenient fixed
datum (a horizontal or a vertical fixed axis). The total
length of the rope is
x+
p r2
+ 2 y + p r1 + b = L …(1)
2
Since L, r2, r1, and b are all constant, so the first and
second time derivatives of the equation (1) give
x + 2 y = 0
⇒
vA + 2vB = 0 …(2)
⇒
v A + 2v B = 0 or x + 2 y = 0 ⇒
aA + 2 aB = 0 …(3)
The velocity and acceleration constraint equations
indicate that, for the coordinates selected, the velocity of A must have a sign which is opposite to that of
the velocity of B, and similarly for the accelerations.
The constraint equations are valid for the motion of
the system in either direction. We emphasize that
vA = x is positive to the left and that vB = y is positive down. The system shown is said to have one
degree of freedom since only one variable, either x
or y, is needed to specify the positions of all parts of
the system.
11/28/2019 7:27:09 PM
6.10 JEE Advanced Physics: Mechanics – I
Two Degrees of Freedom
Consider a system, shown, having two degrees of
freedom. Here the positions of the lower cylinder E
and pulley C depend on the separate specifications
of the two coordinates yA and yB. The lengths of the
cables (measured from DATUM), attached to cylinders A and B can be written, respectively, as
C
A
y A + 2 yD + constant = LA …(1)
yB + yC + ( yC - yD ) + constant = LB …(2)
B
Solution
DATUM
yA
yB
yC
A
B
yD
The centers of the pulleys at A and B are located by
the coordinates yA and yB measured from fixed positions. The total constant length of cable in the pulley
system is
D
L = 3 yB + 2 y A + constants
C
E
and their time derivatives are
yB
yA
C
y A + 2 y D = 0 and y B + 2 y C - y D = 0
yA + 2 yD = 0 and yB + 2 yC - yD = 0
A
Eliminating the terms in y D and yD gives
y A + 2 y B + 4 y C = 0 or vA + 2vB + 4vC = 0
yA + 2 yB + 4 yC = 0 or aA + 2 aB + 4 aC = 0
It is clearly impossible for the signs of all three terms
to be positive simultaneously. So, for example, if both
A and B have downward (positive) velocities, then C
will have an upward (negative) velocity.
Illustration 6
In the pulley configuration shown, cylinder A has a
downward velocity of 0.3 ms -1. Determine the velocity of B.
06_Newtons Laws of Motion_Part 1.indd 10
B
where the constants account for the fixed lengths of
cable in contact with the circumferences of the pulleys, and the constant vertical separation between
the two upper ­left-hand pulleys. Differentiation with
time gives
0 = 3 y B + 2 y A
Substitution of vA = y A = 0.3 ms -1 and vB = y B gives
0 = 3 ( vB ) + 2 ( 0.3 )
⇒
vB = -0.2 ms -1 11/28/2019 7:27:14 PM
Chapter 6: Newton’s Laws of Motion
Illustration 7
The car A is used to pull a load B with the pulley
arrangement shown. If A has a forward velocity vA,
determine an expression for the upward velocity vB
of the load in terms of x.
later on the developed relations can be linked for getting the required parameters. Sometime x and y directional motion or any two directions of the motion are
related by some specific rule, we call such rules as
constraint rules. These rules relate one direction of
motion of an object with some other direction of the
same object or some other object also.
Illustration 8
l
h
6.11
B
A
x
Figure shows a rod of length l resting on a wall and
the floor. Its lower end A is pulled towards left with a
constant velocity u. Find the velocity of the other end
B downward when the rod makes an angle q with the
horizontal.
Solution
B
We designate the position of the car by the coordinate
x and the position of the load by the coordinate y,
both measured from a fixed reference. The total constant length of the cable is
l
L = 2 ( h - y ) + l = 2 ( h - y ) + h2 + x 2
A
v
θ
Solution
h
METHOD I
Here if the distance from the corner to the point A is x
and that up to B is y. The velocity of point A can then
be given by
l
y
B
A
x
Differentiation with time yields
0 = -2 y +
xx
2
2
h +x
Substituting vA = x and vB = y gives
1 xvA
vB =
2 h2 + x 2
dx
dt
and that of B can be given as
v=
dy
dt
Since, we have,
vB =
x2 + y 2 = l2
B
Simple Constraint Motion of Bodies
and Particles in Two Dimensions
Similar to projectile motion, there can be several two
dimensional motions, in which the laws of motion
can be separately applied to x and y directions and
06_Newtons Laws of Motion_Part 1.indd 11
l
v
y
A
x
11/28/2019 7:27:16 PM
6.12 JEE Advanced Physics: Mechanics – I
Differentiating with respect to t, we get
2x
Illustration 9
Figure shows a hemisphere and a supported rod.
Hemisphere is moving in right direction with a uniform velocity v2 and the end of rod which is in contact
with ground is moving in left direction with a velocity v1. Find the rate at which the angle q is changing
in terms of v1, v2, R and q.
dy
dx
+ 2y
=0
dt
dt
xv + yvB = 0
⎛ x⎞
vB = - ⎜ ⎟ v
⎝ y⎠
⇒
⇒
vB = -v cot q {negative sign indicates, y decreasing with time}
METHOD II
In cases when the relation between two points of a
rigid body is required, we can make use of the fact
that in a rigid body the distance between two points
always remains same. Thus the relative velocity of
one point of an object with respect to any other point
of the same object in the direction of line joining them
will always remain zero, as their separation always
remains constant.
Here in above example the distance between the
points A and B of the rod always remains constant,
thus, the two points must have same velocity
components in the direction of their line joining i.e.,
along the length of the rod.
If point B is moving down with velocity vB, its
component along the length of the rod is vB sin q .
Similarly the velocity component of point A along the
length of rod is v cos q . Thus we have.
vB sin q = v cos q
For A
θ
v2
Solution
Here x is the separation between centre of hemisphere
and the end of rod. Rate of change of x is actually
the relative velocity of end of rod and centre of
hemisphere i.e., ( v1 + v2 ). We are required to find the
dq
dx
= v1 + v2
, knowing that
dt
dt
rate of change of q, i.e.,
Since, x = R cosec q
Differentiating with respect to time we get
dx
dq
= - R cosec q cot q
dt
dt
{
∵
d
( cosec q ) = - cosec q cot q
dt
}
dq ( v1 + v2 ) sin q
=
dt
R cos q
2
Illustration 10
vcos(180 – θ )
v
R
θ
x
⇒
vB = v cot q
180 – θ
v1
90 – θ
vBsinθ
vB
For B
In the system shown, if a1, a2 and a3 be the respective
accelerations of 1, 2 and 3, then find a1 in terms of a2
and a3.
Conceptual Note(s)
In such type of problems, when velocity of one part
of a body is given and that of other is required or in
cases, when relation in two velocities is required, we
first find the relation between the two displacements
then differentiate with respect to time.
06_Newtons Laws of Motion_Part 1.indd 12
1
2
3
11/28/2019 7:27:20 PM
Chapter 6: Newton’s Laws of Motion
6.13
Solution
Let the datum pass from the centre of the fixed ­pulley.
Since the points 1, 2, 3 and 4 are movable, so let their
displacements at any instant from the datum be x1, x2,
x3 and x4. We observe that
x1 + x 4 = l1 P
…(1)
(length of first string between 1 and 4 is constant)
and
A
B
C
( x2 - x4 ) + ( x3 - x4 ) = l2 (length of second string between 2 and 3 is constant)
Solution
⇒
Let v be the velocity of block A (upwards) and vr be
the velocity of block B with respect to moving pulley
P (upwards).
Taking upward direction as positive, we have
x2 + x3 - 2x 4 = l2 …(2)
Differentiating twice with respect to time, we get
x1 + x4 = 0 and x2 + x3 - 2x4 = 0
⇒
vA = + v, vB = vr - v and vC = - ( vr + v )
a1 + a4 = 0 …(3)
and a2 + a3 - 2 a4 = 0 …(4)
Now it is given that,
vAC = +300 mms -1
DATUM
x1
x2
x3
4
2
Now, since a4 = - a1, so we get
⇒
2v + vr = 300 mms -1
⇒
vB - vA = -200 mms -1
⇒
-1
vr - 2v = -200 mms …(2)
vr = 50 mms -1 and v = 125 mms -1
{from equation (3)}
So, vA = + v = 125 mms -1 {upwards}
vB = vr - v = -75 mms -1
(in magnitude)
Illustration 11
In the arrangement shown, the three blocks move
with constant velocities. Knowing that the relative
velocity of A with respect to C is 300 mms -1 upwards
and that the relative velocity of B with respect to A
is 200 mms -1 downwards, find the velocity of each
block.
06_Newtons Laws of Motion_Part 1.indd 13
⇒
Solving equations (1) and (2), we get
3
a2 + a3 + 2 a1 = 0
a + a3
a1 = 2
2
vA + vC = 300 mms -1 …(1)
Similarly vBA = -200 mms -1
x4
1
⇒
⇒
vB = 75 mms -1 , downwards
and vC = -vr - v = -175 mms -1
⇒
-1
vC = 175 mms , downwards.
Illustration 12
In the system shown, if a0 be the acceleration of
block 1, then find the acceleration of block 2.
11/28/2019 7:27:29 PM
6.14 JEE Advanced Physics: Mechanics – I
When 1 moves up by x, 3 goes down by x, 4 by 2x
and 2 by 4x, so that total distance moved down by 2
becomes ( x + 2x + 4 x ) = 7 x .
Illustration 13
Determine the relationship which governs the velocities of the four cylinders. Express all velocities as
positive down. How many degrees of freedom are
there?
2
1
Solution
Let the datum pass through the centre of the fixed
pulley. Since the points 1, 2, 3 and 4 are movable,
so let their displacements from the datum be x1, x2,
x3 and x4. We observe